\(\displaystyle\int\limits_1^2 3f(x)\mathrm{\,d}x=3\displaystyle\int\limits_1^2 f(x)\mathrm{\,d}x=3\cdot2=6\).

Ta có \(\displaystyle\int\limits_{2}^{3}2f(x)\mathrm{\,d}x\)= \(2 \displaystyle\int\limits_{2}^{3}f(x)\mathrm{\,d}x=2 \cdot 6=12\).

Ta có \(\displaystyle\int\limits_2^3\left[f(x)+g(x)\right]\mathrm{\,d}x=\int\limits_2^3 f(x)\mathrm{\,d}x+\int\limits_2^3 g(x)\mathrm{\,d}x=3+1=4\).

\(\displaystyle\int\limits_{0}^{3}[f(x)+g(x)]\mathrm{\,d}x=\displaystyle\int\limits_{0}^{3}f(x)\mathrm{\,d}x+\displaystyle\int\limits_{0}^{3}g(x)\mathrm{\,d}x=7\).

\(\displaystyle\int\limits_{0}^{3}[f(x)-g(x)]\mathrm{\,d}x=\displaystyle\int\limits_{0}^{3}f(x)\mathrm{\,d}x-\displaystyle\int\limits_{0}^{3}g(x)\mathrm{\,d}x=3\).

\(\displaystyle\int\limits_{0}^{3} 3 f(x) \mathrm{d} x=3\displaystyle\int\limits_{0}^{3} f(x) \mathrm{d} x=15\).

\(\displaystyle\int\limits_{0}^{3}[2 f(x)-3 g(x)] \mathrm{d} x=2\displaystyle\int\limits_{0}^{3}f(x) \mathrm{d} x-3\displaystyle\int\limits_{0}^{3} g(x)\mathrm{d} x=4\).

Ta có \(\displaystyle\int\limits_{0}^4 f(x) \mathrm{~d}x = \int\limits_{0}^3 f(x) \mathrm{~d}x + \int\limits_{3}^4 f(x) \mathrm{~d}x \), suy ra

\(\displaystyle\int\limits_{0}^3 f(x) \mathrm{~d}x = \int\limits_{0}^4 f(x) \mathrm{~d}x - \int\limits_{3}^4 f(x) \mathrm{~d}x = 4 - 6 = -2. \)

\(\displaystyle\int\limits_1^2 \left[f(x)-g(x)\right]\mathrm{\,d}x=\displaystyle\int\limits_1^2 f(x) \mathrm{\,d}x-\displaystyle\int\limits_1^2 g(x) \mathrm{\,d}x=3-2=1\).

Ta có \(\displaystyle\int\limits_1^2 \left[f(x)+g(x)\right] \mathrm{\,d}x=\displaystyle\int\limits_1^2 f(x) \mathrm{\,d}x+\displaystyle\int\limits_1^2 g(x) \mathrm{\,d}x=3+2=5\).

Ta có \(\displaystyle \int\limits_{0}^{1}\left[f(x)-g(x)\right]\mathrm{\,d}x=\int\limits_{0}^{1}f(x) \mathrm{\,d}x-\int\limits_{0}^{1}g(x)\mathrm{\,d}x=-2-3=-5\).

Ta có \(\displaystyle\int\limits_0^1[f(x)+g(x)]\mathrm{\,d}x=\displaystyle\int\limits_0^1f(x)\mathrm{\,d}x+\displaystyle\int\limits_0^1g(x)\mathrm{\,d}x=3-4=-1\).

Ta có \(\displaystyle\int\limits_{1}^{2} \left[f(x)-g(x)\right]\mathrm{\,d}x=\displaystyle\int\limits_{1}^{2} f(x)\mathrm{\,d}x-\displaystyle\int\limits_{1}^{2} g(x)\mathrm{\,d}x=2-6=-4\).

Ta có \(\displaystyle\int\limits_0^1[f(x)+g(x)]\mathrm{\,d}x=\displaystyle\int\limits_0^1f(x)\mathrm{\,d}x+\displaystyle\int\limits_0^1g(x)\mathrm{\,d}x=3-4=-1\).

\(\displaystyle \int\limits_0^2f(x)\mathrm{\,d}x=\int\limits_0^1f(x)\mathrm{\,d}x+\int\limits_1^2f(x)\mathrm{\,d}x=6.\)

Ta có \( \displaystyle \int\limits_0^2 \left[f(x)+3g(x)\right] \mathrm{\, d}x =\int\limits_0^2 f(x) \mathrm{\, d}x+3\int\limits_0^2 g(x) \mathrm{\, d}x =3+3\cdot 7 =24 \).

Ta có \(\displaystyle\int\limits_{0}^{2}\left[2f(x)-g(x)\right]\mathrm{\, d}x =2\displaystyle\int\limits_{0}^{2}f(x)\mathrm{\, d}x-\displaystyle\int\limits_{0}^{2}g(x)\mathrm{\, d}x=2\cdot3-(-2)=8\).

Ta có

\(\displaystyle\int\limits_{0}^{1}\left[f(x)+2g(x)\right]\mathrm{\,d}x=\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x+2\displaystyle\int\limits_{0}^{1}g(x)\mathrm{\,d}x=3+2(-2)=-1.\)

Ta có \(\displaystyle \int\limits_{-1}^{3} f(x)\mathrm{d}x = \displaystyle \int\limits_{1}^{3} f(x)\mathrm{d}x + \displaystyle \int\limits_{-1}^{1} f(x)\mathrm{d}x \Rightarrow \displaystyle \int\limits_{-1}^{1} f(x)\mathrm{d}x = \displaystyle \int\limits_{-1}^{3} f(x)\mathrm{d}x - \displaystyle \int\limits_{1}^{3} f(x)\mathrm{d}x = -4\).

\(I=\displaystyle \int \limits_a^b f(x) \mathrm{\,d}x=\displaystyle \int \limits_a^c f(x) \mathrm{\,d}x+\displaystyle \int \limits_c^b f(x) \mathrm{\,d}x=\displaystyle \int \limits_a^c f(x) \mathrm{\,d}x-\displaystyle \int \limits_b^c f(x) \mathrm{\,d}x=17-(-11)=28.\)

Ta có \(\displaystyle\int\limits_9^0 g(x)\mathrm{\,d}x=16\Rightarrow \displaystyle\int\limits_0^9 g(x)\mathrm{\,d}x=-16\).

Vậy \(I=2\displaystyle\int\limits_0^9 f(x)\mathrm{\,d}x+3\displaystyle\int\limits_0^9 g(x)\mathrm{\,d}x=2\displaystyle\int\limits_0^9 f(x)\mathrm{\,d}x+3\displaystyle\int\limits_0^9 g(x)\mathrm{\,d}x=2\cdot 37+3\cdot(-16)=26\).

Theo tính chất tích phân

\( \int\limits_{1}^{5}f(x)\mathrm{\, d}x=\int\limits_{1}^{2}f(x)\mathrm{\,d}x+ \int\limits_{2}^{5}f(x)\mathrm{\, d}x=3+(-1)=2.\)

Ta có \(\displaystyle \int \limits_2^7f(x)\mathrm{d}x=\displaystyle \int \limits_2^5f(x)\mathrm{d}x+\displaystyle \int \limits_5^7f(x)\mathrm{d}x=3+9=12.\)

Ta có \(\displaystyle \int\limits_1^3 f(x) \mathrm{\,d}x = \displaystyle \int\limits_1^2 f(x) \mathrm{\,d}x + \displaystyle \int\limits_2^3 f(x) \mathrm{\,d}x = 3+4=7\).

Ta có \(K=2\displaystyle\int\limits_1^3 f(x)\mathrm{\,d}x-3\displaystyle\int\limits_1^3 g(x)\mathrm{\,d}x=2\cdot9-3\cdot(-5)=33\).

Ta có

\(\displaystyle\int\limits_{5}^{2}\left[2-4f(x)\right]\mathrm{\,d} x=\int\limits_{5}^{2}2\mathrm{\,d} x-4\int\limits_{5}^{2}f(x)\mathrm{\,d} x=2(2-5)-4\cdot(-10)=34.\)

\(I=\displaystyle\int\limits_a^b f(x) \mathrm{\,d}x = F(x)\Big|_a^b =F(b)-F(a)= 3-(-2)=5 .\)

Ta có \( \displaystyle\int\limits_{-2}^3 f(x) \mathrm{~d}x = -10 \), suy ra \(F(3) - F(-2) = -10\). \\ Vậy \(F(-2) = F(3) + 10 = -8 + 10 =2\).

Ta có \(\displaystyle\int\limits_1^3 f(x) \mathrm{d} x=\displaystyle\int\limits_1^2 f(x) \mathrm{d} x+\displaystyle\int\limits_2^3 f(x) \mathrm{d} x=\frac{1}{2}+\frac{3}{2}=2\).

\(\displaystyle\int\limits_1^3[2 f(x)+g(x)] \mathrm{d} x=2 \displaystyle\int\limits_1^3 f(x) \mathrm{d} x+\displaystyle\int\limits_1^3 g(x) \mathrm{d} x=2\cdot 2-1=3.\)

\(\displaystyle\int\limits_1^3[5 f(x)-4] \mathrm{d} x=5 \displaystyle\int\limits_1^3 f(x) \mathrm{d} x-\displaystyle\int\limits_1^3 4 \mathrm{~d} x=5\cdot 2-(4 x)\bigg|_1 ^3=10-8=2\).

Ta có \(\displaystyle\int\limits_0^3 f'(x) \mathrm{\,d}x=9 \Leftrightarrow f(x) \bigg|_0^3=9\Leftrightarrow f(3)-f(0)=9\)

\(\Leftrightarrow f(3)=9+f(0)=9+1=10\).

Vậy \(f(3)=10\).

Ta có \(I=\displaystyle\int\limits_1^2 f'(x)\mathrm{\,d}x =f(x)\bigg|_1^2=f(2)-f(1)=2-1=1\).

\(\displaystyle \int\limits_{a}^{b}f'(x) \mathrm{\, d}x= 3\sqrt{5} \Rightarrow f(b)-f(a)=3\sqrt{5} \Rightarrow f(a) = f(b) - 3\sqrt{5}= \sqrt{5}(\sqrt{5}-3)\).

Ta có \(\displaystyle \int_{0}^{1} f(x) \mathrm{\, d}x=F(1)-F(0)=3\).

Ta có \(I=\displaystyle \int \limits_{-5}^{3}\left[7f(x)-x\right] \mathrm{d} x = 7F(x)\bigg|_{-5}^3 - \displaystyle\frac{x^2}{2}\bigg|_{-5}^3 = 2\).

\(I=\displaystyle \int\limits_{2}^{4} f''(x)\mathrm{\,d}x=f'(x)\big|_2^4=f'(4)-f'(2)=5-1=4\).

\(I=\displaystyle\int\limits_1^3 f'(x)\mathrm{\,d}x=f(x) \big|_1^3=f(3)-f(1)=9-2=7.\)

\(I=\displaystyle\int\limits_a^b [2f(x)-3g(x)] \mathrm{\,d}x=2\int\limits_a^b f(x) \mathrm{\,d}x-3\int\limits_a^b g(x) \mathrm{\,d}x=2\cdot (-2)-3\cdot 3=-13\).

Ta có:

\begin{eqnarray*}P&=&\displaystyle\int\limits_{-2}^1f(x){\mathrm{d}x}+\displaystyle\int\limits_3^5f(x){\mathrm{d}x} \\ &=&\displaystyle\int\limits_{-2}^1f(x){\mathrm{d}x}+\displaystyle\int\limits_1^3f(x){\mathrm{d}x}+\displaystyle\int\limits_3^5f(x){\mathrm{d}x}-\displaystyle\int\limits_1^3f(x){\mathrm{d}x}\\ &=&\displaystyle\int\limits_{-2}^5f(x){\mathrm{d}x}-\displaystyle\int\limits_1^3f(x){\mathrm{d}x}\\ &=&8-(-3)=11.\end{eqnarray*}

\(\displaystyle \int \limits_{1}^{4}f'(x)\mathrm{\,d}x=f(4)-f(1) =8-15= -7.\)

Ta có \(L=\displaystyle\int\limits_0^3 \left[2f(x)-g(x)\right] \mathrm{\,d}x=2\displaystyle\int\limits_0^3 f(x) \mathrm{\,d}x-\displaystyle\int\limits_0^3 g(x) \mathrm{\,d}x=2\cdot 2-3=1\).

\(\displaystyle\int\limits_0^1 [f(x)+g(x) ]\mathrm{\,d}x= \displaystyle\int\limits_0^1 f(x) \mathrm{\,d}x + \displaystyle\int\limits_0^1 g(x) \mathrm{\,d}x=-1\).

\(\displaystyle\int\limits_{0}^{2} f'\left(x\right)\mathrm{\,d}x = f\left(x\right) \bigg \vert_{0}^2 = f\left(2\right) - f\left(0\right) = 5 \Leftrightarrow f\left(2\right) = 4\).

Ta có \(\displaystyle\int\limits_{2}^{5}\left[f(x)-2g(x)\right]\mathrm{\,d}x=\displaystyle\int\limits_{2}^{5}f(x)\mathrm{\,d}x-2\displaystyle\int\limits_{2}^{5}g(x)\mathrm{\,d}x=3-2\cdot 9=-15\).

Ta có \(\displaystyle\int\limits_{0}^{1}2f(x)\mathrm{d}x+5=2\displaystyle\int\limits_{0}^{1}f(x)\mathrm{d}x+5=11\).

Ta có \(\displaystyle \int \limits_2^5 \left[ f(x)+ g(x)\right]\,\mathrm{d}x=\int \limits_2^5 f(x)\,\mathrm{d}x + \int \limits_2^5 g(x) \, \mathrm{d}x = 3 + 9 = 12\).

Ta có \(\displaystyle\int\limits_{0}^{2}\left[ f(x)-g(x) \right] \mathrm{\,d}x = \displaystyle\int\limits_{0}^{2} f(x)\mathrm{\,d}x - \displaystyle\int\limits_{0}^{2}g(x)\mathrm{\,d}x = 5\).

Ta có \(\displaystyle\int\limits_{0}^{2}f(x)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x+\displaystyle\int\limits_{1}^{2}f(x)\mathrm{\,d}x=1+2=3.\)

Ta có \(\displaystyle \int \limits_0^3 [x - 3f(x)] \mathrm{d}x = \displaystyle \int \limits_0^3 x \mathrm{d}x - 3\displaystyle \int \limits_0^3 f(x) \mathrm{d}x = \displaystyle\frac{1}{2}x^2 \Big|^3_0 - 6 = \displaystyle\frac{9}{2} - 6 = -\displaystyle\frac{3}{2}\).

Ta có \(\displaystyle\int\limits_1^5[4f(x)-g(x)]\mathrm{\,d}x=4\displaystyle\int\limits_1^5f(x)\mathrm{\,d}x-\displaystyle\int\limits_1^5g(x)\mathrm{\,d}x=4\cdot 6-8=16\).

Ta có \(\displaystyle\int\limits_0^2 f'(x)\mathrm{\,d}x=4\Leftrightarrow f(x) \Big|_0^2=4\Leftrightarrow f(2)-f(0)=4 \Leftrightarrow f(0)-f(2)=-4\).

Đặt \(I_1=\displaystyle\int\limits_{-1}^{5}f(x)\mathrm{d}x\), \(I_2=\displaystyle\int\limits_{-1}^{5}g(x)\mathrm{d}x\).

Ta có

\begin{align*}&\displaystyle\int\limits_{-1}^{5}\left[2f(x)+3g(x)\right]\mathrm{\,d}x=-5\Leftrightarrow 2\displaystyle\int\limits_{-1}^{5}f(x)\mathrm{\,d}x+3\displaystyle\int\limits_{-1}^{5}g(x)\mathrm{\,d}x=-5\Rightarrow 2I_1+3I_2=-5.\\ &\displaystyle\int\limits_{-1}^{5}\left[3f(x)-5g(x)\right]\mathrm{\,d}x=21\Leftrightarrow 3\displaystyle\int\limits_{-1}^{5}f(x)\mathrm{\,d}x-5\displaystyle\int\limits_{-1}^{5}g(x)\mathrm{\,d}x=21\Rightarrow 3I_1-5I_2=21.\end{align*}

Từ đó suy ra \(I_1=2\), \(I_2=-3\).

Vậy \(\displaystyle\int\limits_{-1}^{5}\left[f(x)+g(x)\right]\mathrm{\,d}x=\displaystyle\int\limits_{-1}^{5}f(x)\mathrm{\,d}x+\displaystyle\int\limits_{-1}^{5}g(x)\mathrm{\,d}x=I_1+I_2=-1\).

Ta có \(\displaystyle\int\limits_0^3 f'(x)\mathrm{\,d}x=f(x)\bigg|_0^3 \Leftrightarrow 9=f(3)-f(0) \Leftrightarrow f(3)=9+f(0)=10\).

Ta có

\(I=\displaystyle\int\limits_0^3 f(x)\mathrm{\,d}x = \displaystyle\int\limits_0^2 f(x)\mathrm{\,d}x +\displaystyle\int\limits_2^3 f(x)\mathrm{\,d}x =1+4=5\).

Ta có \(I=\displaystyle\int\limits_0^4 f(x)\mathrm{\,d}x = \displaystyle\int\limits_0^2 f(x)\mathrm{\,d}x + \displaystyle\int\limits_2^4 f(x)\mathrm{\,d}x = 9 + 4 = 13\).

\(\int\limits_{-2}^2 f(x) \mathrm{\,d}x=\int\limits_{-2}^0 f(x) \mathrm{\,d}x+\int\limits_{0}^2 f(x) \mathrm{\,d}x=3+4=7\).

Ta có: \(\displaystyle\int\limits_1^2{\left[2+f\left(x\right)\right]}\mathrm{d}x = \left(2x + x^2\right)\bigg|_{1}^{2}=5\).

Ta có

\(\displaystyle\int\limits_{1}^{2} [2 + f(x)]\mathrm{\,d}x = 2\displaystyle\int\limits_{1}^{2}\mathrm{\,d}x + \displaystyle\int\limits_{1}^{2}f(x)\mathrm{\,d}x =2x\Big|^2_1+F(x)\Big|^2_1= 2x\Big|^2_1 + x^3\Big|^2_1 = 9.\)

Ta có \(\displaystyle\int\limits_1^3 \left[1+f(x)\right] \mathrm{\,d}x=\displaystyle\int\limits_1^3 1 \mathrm{\,d}x+\displaystyle\int\limits_1^3 f(x) \mathrm{\,d}x=2+x^3\Big|_1^3=2+(27-1)=28\).

Ta có \(\displaystyle\int\limits_{1}^{3} \left[1+f(x)\right] \mathrm{\,d}x=x\Big|_1^3+\displaystyle\int\limits_{1}^{3} f(x) \mathrm{\,d}x=2+F(x)\Big|_1^3=2+x^2\Big|_1^3=10\).

Ta có \( \displaystyle \int\limits_0^1\left[f(x)+2x \right] \mathrm{\,d}x=\displaystyle \int\limits_0^1 f(x) \mathrm{\,d}x+\displaystyle \int\limits_0^1 2x \mathrm{\,d}x=\displaystyle \int\limits_0^1 f(x) \mathrm{\,d}x+1=2\).

Suy ra \( \displaystyle \int\limits_0^1 f(x) \mathrm{\,d}x =1\).

}

Ta có \(\displaystyle\int\limits_0^1[f(x)+2x] \mathrm{\,d}x=\displaystyle\int\limits_0^1f(x)\mathrm{\,d}x+\displaystyle\int\limits_0^12x \mathrm{\,d}x=3\).

Suy ra \(\displaystyle\int\limits_0^1f(x)\mathrm{\,d}x=3-\displaystyle\int\limits_0^12x \mathrm{\,d}x=3-\left.x^2\right|_0^1=3-(1-0)=2\).

Ta có,

\begin{eqnarray*}&&\int_0^1 \left(f(x)+2x \right)\mathrm{d}x=4\\ &\Leftrightarrow&\int_0^1f(x)\mathrm{d}x +\int_0^1 2x \mathrm{d}x=4\\ &\Leftrightarrow&\int_0^1f(x)\mathrm{d}x+2=4\\ &\Leftrightarrow&\int_0^1f(x)\mathrm{d}x=2.\end{eqnarray*}

Ta có

\begin{eqnarray*}\displaystyle\int\limits_0^1\left[f(x)+2x\right]\mathrm{\,d}x=5 &\Leftrightarrow& \displaystyle\int\limits_0^1f(x)\mathrm{\,d}x+\displaystyle\int\limits_0^1 2x\mathrm{\,d}x=5\\ &\Leftrightarrow&\displaystyle\int\limits_0^1f(x)\mathrm{\,d}x+x^2\Big|_0^1=5\\ &\Leftrightarrow&\displaystyle\int\limits_0^1f(x)\mathrm{\,d}x=5-1=4.\end{eqnarray*}

Ta có \(-1=\displaystyle\int\limits_{1}^{2}[kx-f(x)]\mathrm{\,d}x=\displaystyle\frac{kx^{2}}{2}\Big |_{1}^{2}-4\Rightarrow\displaystyle\frac{3k}{2}=3\Rightarrow k=2\).

\(I= \displaystyle \int_{0}^{2} [2x+f(x)-2g(x)] \mathrm{\, d}x=\displaystyle \int_{0}^{2} 2x \mathrm{\, d}x +\displaystyle \int_{0}^{2} f(x)\mathrm{\, d}x -2\displaystyle \int_{0}^{2} g(x) \mathrm{\, d}x = x^2\big|_0^2+3-2\cdot (-2) = 11\).

Ta có

\begin{equation*}I=\displaystyle \int\limits_{-1}^2 \left[x+2f(x)-3g(x)\right]\mathrm{d}x=\displaystyle\frac{x^2}{2}\bigg|_{-1}^2 +2\displaystyle \int\limits_{-1}^2 f(x)\mathrm{d}x-3\displaystyle \int\limits_{-1}^2 g(x)\mathrm{d}x=\displaystyle\frac{3}{2}+4+3=\displaystyle\frac{17}{2}.\end{equation*}

Ta có

\( I=\displaystyle\int\limits_{-1}^{3}[2f(x)-x]\mathrm{\,d}x=\left.\left(2F(x)-\displaystyle\frac{x^2}{2}\right)\right |_{-1}^{3}=2[F(3)-F(-1)]-4=3. \)

Ta có

\begin{align*}I = \int \limits_0^1 [3\mathrm{e}^x -f(x)]\ \mathrm{d}x = 3\int \limits_0^1 \mathrm{e}^x \ \mathrm{d}x - \int \limits_0^1 f(x)\ \mathrm{d}x = 3 (\mathrm{e} -1) -7 = 3\mathrm{e} -10.\end{align*}

Ta có

\(\displaystyle\int\limits_{2}^{4}[2 f(x)+x-1] \mathrm{\,d}x=2 \displaystyle\int\limits_{2}^{4} f(x) \mathrm{\,d}x+\displaystyle\int\limits_{2}^{4}(x-1) \mathrm{\,d}x=2\cdot 3+4=10\).

Ta có \(\displaystyle \int\limits_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{\left[ \cos 2x+2f(x) \right]\mathrm{d}x}=\displaystyle \int\limits_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{ \cos 2x \mathrm{d}x}+2\displaystyle \int\limits_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{f(x) \mathrm{d}x}=\displaystyle\frac{1}{2} \sin{2x}\Big|_{\frac{\pi}{4}}^{\frac{\pi}{4}}+2I=\displaystyle\frac{1}{2}(1+1)+2I=1+2I.\)

Theo đề ta có \(1+2I=5 \Leftrightarrow I=2\).

Ta có \( I=\displaystyle\int\limits_{-1}^26x^2\mathrm{\,d}x =\left.\left(2x^3\right)\right|_{-1}^2=2\left[2^3-(-1)^3\right]=18\).

\(\displaystyle\int\limits_1^2 x^4 \mathrm{\,d}x=\left.\displaystyle\frac{1}{5} \cdot x^{5}\right|_1 ^2=\displaystyle\frac{1}{5}\left(2^5-1\right)=\displaystyle\frac{31}{5}\).

\(\begin{aligned}\int\limits_0^1\left(4 x^3+3 x^2-2\right) \mathrm{d} x &=\int\limits_0^1 4 x^3 \mathrm{~d} x+\int\limits_0^1 3 x^2 \mathrm{~d} x-\int\limits_0^1 2 \mathrm{~d} x \\& =\left.x^4\right|_0 ^1+\left.x^3\right|_0 ^1-\left.2 x\right|_0 ^1 \\ & =\left(1^4-0^4\right)+\left(1^3-0^3\right)-(2.1-2.0)=0.\end{aligned}\)

\(\displaystyle\int\limits_2^3 3 x^2 \mathrm{~d} x=x^3\bigg|_2 ^3=3^3-2^3=27-8=19\).

\(\displaystyle\int_{0}^{1}(3x+1)(x+3)\mathrm{\,d}x=\displaystyle\int_{0}^{1}(3x^2+10x+3)\mathrm{\,d}x=\left. {{x}^{3}+5x^{2}+3x} \right|_{0}^{1}=9\).

\(\displaystyle\int\limits_{0}^{3}\left(9x^2-6x+1\right)\mathrm{\,d}x=\displaystyle\int\limits_0^3 9 x^2 \mathrm{\,d}x-\displaystyle\int\limits_0^3 6 x \mathrm{\,d}x+\displaystyle\int\limits_0^3 1 \mathrm{\,d}x=3x^3\Big|_0^3-3x^2\Big|_0^3+x\Big|_0^3=57.\)

Ta có:

\(\displaystyle\int\limits_0^1\left(x^2+x\right) \mathrm{d} x=\displaystyle\int\limits_0^1 x^2 \mathrm{~d} x+\displaystyle\int\limits_0^1 x \mathrm{~d} x=\left.\displaystyle\frac{x^3}{3}\right|_0 ^1+\left.\displaystyle\frac{x^2}{2}\right|_0 ^1=\left(\displaystyle\frac{1}{3}-0\right)+\left(\displaystyle\frac{1}{2}-0\right)=\displaystyle\frac{5}{6}\).

\(\displaystyle\int\limits_0^1\left(x^6-4 x^3+3 x^2\right) \mathrm{d} x =\left. \left(\displaystyle\frac{x^7}{7}-x^4+x^3\right)\right|_{0}^1 = \left(\displaystyle\frac{1}{7} - 1 + 1\right) - 0 = \displaystyle\frac{1}{7}\).

Đặt \(u=x+2\) ta có \(\mathrm{d}x\,=\mathrm{d}u\,\). Khi \(x=-1\Rightarrow u=1\). \(x=1\Rightarrow u=3\).

Khi đó \(\displaystyle\int\limits_{-1}^1(x+2)^3\mathrm{d}x\,=\displaystyle\int\limits_1^3u^3\mathrm{d}u\,=\displaystyle\frac{u^4}{4}\Big|_1^3=\displaystyle\frac{3^4-1^4}{4}=20\).

\begin{eqnarray*}& \displaystyle\int\limits_{-1}^2\left(3 x^2-8 x\right) \mathrm{\,d} x&= \displaystyle\int\limits_{-1}^2 3 x^2 \mathrm{\,d} x-\displaystyle\int\limits_{-1}^2 8 x \mathrm{\,d} x=3 \displaystyle\int\limits_{-1}^2 x^2 \mathrm{\,d} x-8 \displaystyle\int\limits_{-1}^2 x \mathrm{\,d} x\\ &&=\left.3 \cdot\left(\displaystyle\frac{x^3}{3}\right)\right|_{-1} ^2-\left.8 \cdot\left(\displaystyle\frac{x^2}{2}\right)\right|_{-1} ^2=\left[2^3-(-1)^3\right]-4\left[2^2-(-1)^2\right]=-3.\end{eqnarray*}

\(\displaystyle\int\limits_1^2 x^2 \mathrm{\,d} x=\left.\displaystyle\frac{x^3}{3}\right|_1^2=\displaystyle\frac{1}{3}\left(2^3-1^3\right)=\displaystyle\frac{7}{3}\).

\(\displaystyle\int\limits_{1}^{2} (x+3)^2 \mathrm{\,d}x =\displaystyle\int\limits_{1}^{2} (x+3)^2 \mathrm{\,d}(x+3)=\displaystyle\frac{(x+3)^3}{3}\Big |_1^2=\displaystyle\frac{61}{3}\).

Ta có \(\displaystyle\int\limits_0^2 \left(x^2-3x \right)\mathrm{\,d}x=\left(\displaystyle\frac{x^3}{3}-\displaystyle\frac{3x^2}{2}\right)\bigg|_0^2=\displaystyle\frac{-10}{3}\).

Ta có \(\displaystyle\int\limits_0^1 x(x+1) \mathrm{\,d}x=\displaystyle\int\limits_0^1 (x^2+x) \mathrm{\,d}x=\left(\displaystyle\frac{x^3}{3}+\displaystyle\frac{x^2}{2}\right) \bigg|_0^1=\displaystyle\frac{5}{6}\).

Ta có \(\displaystyle\int\limits_{0}^{3}(2x+1)\mathrm{\,d}x = (x^2+x)\Big|_{0}^{3} = 12.\)

\(I=\displaystyle\int\limits_1^3 \left(4x^3+3x \right)\mathrm{\,d}x=\left. \left(x^4+\displaystyle\frac{3}{2}x^2 \right) \right|_1^3=92\).

Ta có \(I=(x^2+x)\bigg|_{-1}^{0}=0\).

\(\begin{aligned} \int\limits_1^4 \frac{1}{2 \sqrt{x}} \mathrm{~d} x & =\frac{1}{2} \int\limits_1^4 x^{-\frac{1}{2}} \mathrm{~d} x \\ & =\left.x^{\frac{1}{2}}\right|_1 ^4=\left.\sqrt{x}\right|_1 ^4=\sqrt{4}-\sqrt{1}=1.\end{aligned}\)

\(\int\limits_{1}^3 x^{\sqrt{2}} \mathrm{d}x = \left. \displaystyle\frac{x^{\sqrt{2}+1}}{\sqrt{2}+1} \right|_{1}^3 = \displaystyle\frac{3^{\sqrt{2}+1}-1}{\sqrt{2}+1}.\)

\(\displaystyle\int\limits_1^2 \frac{1}{x^4} \mathrm{~d} x = \left.\displaystyle\frac{x^{-4+1}}{-4+1}\right|_1^2 = \displaystyle\frac{2^{-3}}{-3} - \displaystyle\frac{1^{-3}}{-3} = \displaystyle\frac{7}{24} \).

\(\displaystyle\int\limits_1^4 \frac{1}{x \sqrt{x}} \mathrm{~d} x = \int\limits_1^4 \frac{1}{x^{\frac{3}{2}}} \mathrm{~d} x = \left. \displaystyle\frac{x^{-\frac{3}{2}+1}}{-\frac{3}{2}+1} \right|_1^4 = \displaystyle\frac{4^{-\frac{1}{2}}}{-\frac{1}{2}} - \displaystyle\frac{1^{-\frac{1}{2}}}{-\frac{1}{2}} = 1\).

\(\displaystyle\int\limits_1^2\displaystyle\frac2{x^2}\mathrm{d}x\,=2\displaystyle\int\limits_1^2x^{-2}\mathrm{d}x\,=-\displaystyle\frac2x\Big|_1^2=-\displaystyle\frac21+\displaystyle\frac22=-1\).

\(\displaystyle\int\limits_1^4x^2\sqrt x\mathrm{d}x\,=\displaystyle\int\limits_1^4x^{\tfrac52}\mathrm{d}x\,=\displaystyle\frac27x^{\tfrac72}\Big|_1^4=\displaystyle\frac27(4^{\tfrac72}-1^{\tfrac72})=\displaystyle\frac27\cdot127=\displaystyle\frac{254}{7}.\)

\(\displaystyle\int\limits_1^2 \displaystyle\frac{1}{4 x^2} \mathrm{\,d} x=\displaystyle\frac{1}{4} \displaystyle\int\limits_1^2 x^{-2} \mathrm{\,d} x=\left.\displaystyle\frac{1}{4} \cdot \displaystyle\frac{1}{-1} x^{-1}\right|_1 ^2=-\left.\displaystyle\frac{1}{4} \cdot \displaystyle\frac{1}{x}\right|_1 ^2=-\displaystyle\frac{1}{4}\left(\displaystyle\frac{1}{2}-1\right)=\displaystyle\frac{1}{8}\).

Ta có \(\displaystyle \int\limits_0^3 \mathrm{\,d}x=x\Big|_0^3=3\).

\(\displaystyle\int\limits_1^2 \displaystyle\frac{1}{\sqrt{x}} \mathrm{\,d}x=\displaystyle\int\limits_1^2 \displaystyle\frac{2}{2\sqrt{x}} \mathrm{\,d}x=\left.2\sqrt{x}\right|_1 ^2=2\left(\sqrt{2}-1\right)\).

\(\displaystyle\int\limits_1^4\left(x^2+6 \sqrt{x}\right) \mathrm{d} x=\displaystyle\int\limits_1^4 x^2 \mathrm{~d} x+6 \displaystyle\int\limits_1^4 x^{\tfrac{1}{2}} \mathrm{~d} x=\displaystyle\frac{x^3}{3}\bigg|_1 ^4+6\left(\displaystyle\frac{2}{3} x^{\tfrac{3}{2}}\right)\bigg|_1 ^4\) \(=\displaystyle\frac{4^3-1}{3}+4\left(2^3-1\right)=49\).

\(\displaystyle\int\limits_{1}^{4}\left(x^{3}+3 \sqrt{x}\right) \mathrm{d} x =\displaystyle\int\limits_{1}^{4} x^{3} \mathrm{~d} x+3 \displaystyle\int\limits_{1}^{4} \sqrt{x} \mathrm{~d} x=\left.\displaystyle\frac{x^{4}}{4}\right|_{1} ^{4}+\left.3 \cdot \displaystyle\frac{x^{\displaystyle\frac{3}{2}}}{\displaystyle\frac{3}{2}}\right|_{1} ^{4} =\displaystyle\frac{311}{4}.\)

\(\displaystyle \int\limits_{1}^{16} \displaystyle\frac{x-1}{\sqrt{x}}\mathrm{\,d}x =\displaystyle \int\limits_{1}^{16} \left( \sqrt{x} - \displaystyle\frac{1}{\sqrt{x}}\right) = \left(\displaystyle\frac{2}{3}x\sqrt{x} -2\sqrt{x}\right)\bigg|^{16}_{1} = 36\).

\(\displaystyle\int\limits_{-2}^4(x+1)(x-1) \mathrm{\,d}x=\displaystyle\int\limits_{-2}^4(x^2-1) \mathrm{\,d}x=18\).

\(\displaystyle\int\limits_1^2 \displaystyle\frac{x^2-2 x+1}{x} \mathrm{\,d}x=\displaystyle\int\limits_1^2\left(x-2+\frac{1}{x}\right) \mathrm{\,d} x=\left.\displaystyle\frac{1}{2}x^2\right|_1^2-\left.2x\right|_1^2+\left.\ln x\right|_1^2=-\displaystyle\frac{1}{2}+\ln 2\).

\(\displaystyle \int\limits_{1}^{4} \left(x^3 -2\sqrt{x}\right)\mathrm{\,d}x = \left(\displaystyle\frac{x^4}{4} - \displaystyle\frac{4}{3}x\sqrt{x}\right) \bigg|^{4}_{1} = \displaystyle\frac{653}{12}\).

Hàm số \(f(x)\) liên tục trên các khoảng \((-\infty . 1)\) và \((1 .+\infty)\).

Ta có \(f(1)=1, \lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1} x^2=1, \lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1}(2 x-1)=1\).

Vi \(\lim_{x \rightarrow 1^*}f(x)=\lim _{x \rightarrow 1^*} f(x)=f(1)\) nên hàm số \(f(x)\) liên tục tại \(x=1\).

Do đó hàm số \(f(x)\) liên tục trèn \(\mathbb{R}\).

\begin{eqnarray*}\displaystyle\int\limits_0^3 f(x) \mathrm{d} x&=&\displaystyle\int\limits_0^1 f(x) \mathrm{d} x+\displaystyle\int\limits_1^3 f(x) \mathrm{d} x\\ &=&\displaystyle\int\limits_0^1 x^2 \mathrm{~d} x+\displaystyle\int\limits_1^3(2 x-1) \mathrm{d} x\\ & =&\displaystyle\frac{x^3}{3}\bigg|_0 ^1+\left(x^2-x\right)\bigg|_1 ^3\\ &=&\displaystyle\frac{1}{3}+6=\displaystyle\frac{19}{3}.\end{eqnarray*}

Ta có \(\displaystyle\int\limits_{-1}^1|x| \mathrm{d} x=\displaystyle\int\limits_{-1}^0|x| \mathrm{d} x+\displaystyle\int\limits_0^1|x| \mathrm{d} x=\displaystyle\int\limits_{-1}^0(-x) \mathrm{d} x+\displaystyle\int\limits_0^1 x \mathrm{~d} x\) \(=-\left.\displaystyle\frac{x^2}{2}\right|_{-1} ^0+\left.\displaystyle\frac{x^2}{2}\right|_0 ^1=\left[0-\left(-\displaystyle\frac{1}{2}\right)\right]+\left(\displaystyle\frac{1}{2}-0\right)=\displaystyle\frac{1}{2}+\displaystyle\frac{1}{2}=1\).

Ta có \(\left|x^2-2 x\right|=\begin{cases}-\left(x^2-2 x\right),\quad 0 \leq x \leq 2\\ x^2-2 x,\quad\quad\quad\quad 2 < x \leq 3.\end{cases}\)

Do đó

\(\begin{aligned}\displaystyle\int\limits_0^3\left|x^2-2 x\right|\mathrm{\,d}x &=\displaystyle\int\limits_0^2\left|x^2-2 x\right|\mathrm{\,d}x+\displaystyle\int\limits_2^3\left|x^2-2 x\right|\mathrm{\,d}x \\ &=\displaystyle\int\limits_0^2\left(-x^2+2 x\right) \mathrm{\,d}x+\displaystyle\int\limits_2^3\left(x^2-2 x\right) \mathrm{\,d}x=\left.\left(-\displaystyle\frac{x^3}{3}+x^2\right)\right|_0 ^2+\left.\left(\displaystyle\frac{x^3}{3}-x^2\right)\right|_2 ^3=\displaystyle\frac{8}{3}.\end{aligned}\)

\(\displaystyle\int\limits_{-2}^1|2 x+2|\mathrm{\,d}x=\displaystyle\int\limits_{-2}^{-1}\left(-2x-2\right) \mathrm{\,d}x+\displaystyle\int\limits_{-1}^1\left(2x+2\right) \mathrm{\,d}x=\left.\left(-x^2-2x\right)\right|_{-2}^{-1}+\left.\left(x^2+2x\right)\right|_{-1}^{1}=5\).

\(\displaystyle\int\limits_0^4\left|x^2-4\right|\mathrm{\,d}x=\displaystyle\int\limits_0^2\left(4-x^2\right) \mathrm{\,d}x+\displaystyle\int\limits_2^4\left(x^2-4\right) \mathrm{\,d}x=\left.\left(4x-\displaystyle\frac{x^3}{3}\right)\right|_0 ^2+\left.\left(\displaystyle\frac{x^3}{3}-4x\right)\right|_2^4=16\).

\begin{eqnarray*}\displaystyle\int\limits_{-2}^3|x-2| \mathrm{d} x&=&\displaystyle\int\limits_{-2}^2|x-2| \mathrm{d} x+\displaystyle\int\limits_2^3|x-2| \mathrm{d} x\\ &=&-\displaystyle\int\limits_{-2}^2(x-2) \mathrm{d} x+\displaystyle\int\limits_2^3(x-2) \mathrm{d} x\\ &=&-\left(\displaystyle\frac{x^2}{2}-2 x\right)\bigg|_{-2} ^2+\left(\displaystyle\frac{x^2}{2}-2 x\right)\bigg|_2 ^3\\ &=&\frac{17}{2}.\end{eqnarray*}

\(\displaystyle\int\limits_{-1}^{2}|2 x+1|\mathrm{\,d}x=\displaystyle\int\limits_{-1}^{-\frac{1}{2}}(2x+1)\mathrm{\,d}x+\displaystyle\int\limits_{-\frac{1}{2}}^{2}(2x+1)\mathrm{\,d}x=\displaystyle\frac{13}{2}\).

Ta có

\(\begin{aligned}\displaystyle\int\limits_{0}^{3}|x-2| \mathrm{d} x & =\displaystyle\int\limits_{0}^{2}|x-2| \mathrm{d} x+\displaystyle\int\limits_{2}^{3}|x-2| \mathrm{d} x=\displaystyle\int\limits_{0}^{2}(2-x) \mathrm{d} x+\displaystyle\int\limits_{2}^{3}(x-2) \mathrm{d} x \\ & =\left.\left(2 x-\displaystyle\frac{x^{2}}{2}\right)\right|_{0} ^{2}+\left.\left(\displaystyle\frac{x^{2}}{2}-2 x\right)\right|_{2} ^{3}=[(4-2)-0]+\left[\left(\displaystyle\frac{9}{2}-6\right)-(2-4)\right]=\displaystyle\frac{5}{2}.\end{aligned}\)

Ta có

\(\displaystyle\int\limits_{0}^{3}|2 x-3|\mathrm{\,d}x=\displaystyle\int\limits_{0}^{\frac{3}{2}}|2 x-3|\mathrm{\,d}x+\displaystyle\int\limits_{\frac{3}{2}}^3|2 x-3|\mathrm{\,d}x=\displaystyle\int\limits_{0}^{\frac{3}{2}}\left(2x-3\right)\mathrm{\,d}x+\displaystyle\int\limits_{\frac{3}{2}}^3\left(2x-3\right)\mathrm{\,d}x=\displaystyle\frac{9}{2}.\)

Ta có \(I=\displaystyle\int\limits_{0}^{\tfrac{\pi}{3}}\sin x\mathrm{\,d}x=-\cos x\Big|_0^{\tfrac{\pi}{3}}=\displaystyle\frac{1}{2}\).

Ta có \(\displaystyle I = \int\limits_0^{\tfrac{\pi}{2}} \cos x\mathrm{\,d}x = \sin x \bigg|_0^{\tfrac{\pi}{2}} =1\).

Ta có: \(\displaystyle\int\limits_0^{\tfrac{\pi}{4}} \sin{x} \mathrm{\,d}x=-\cos{x}\Big|_0^{\tfrac{\pi}{4}}=-\displaystyle\frac{\sqrt{2}}{2}+1=\displaystyle\frac{2-\sqrt{2}}{2}\).

\(\displaystyle 9+\int\limits_{0}^{1}{(2m^{2}x-6m)\mathrm{\,d}x}=0\Leftrightarrow 9+\left[m^2x^2-6mx\right]\Big|_0^1=0\Leftrightarrow m^2-6m+9=0\Leftrightarrow m=3.\)

Ta có \(\displaystyle \int\limits_0^{\tfrac{\pi}{2}} \left(4x-1+\cos x\right)\mathrm{\,d}x=\left(2x^2-x+\sin x\right) \Big|_0^{\tfrac{\pi}{2}}=\pi\left(\displaystyle\frac{\pi}{2}-\displaystyle\frac{1}{2}\right)+1\).

Vậy \(a=2, b=2, c=1\), suy ra \(a-b+c=1\).

\(\displaystyle \int\limits_{0}^{\frac{\pi}{2}} (\cos x - \sin x)\mathrm{\,d}x= \left( \sin x + \cos x\right) \bigg|^{\displaystyle\frac{\pi}{2}}_{0} = 0\).

\(\displaystyle \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{6}} \displaystyle\frac{\mathrm{\,d}x}{\sin^2 x} = -\cot x \bigg|^{\frac{\pi}{4}}_{\frac{\pi}{6}} = -1 +\sqrt{3}\).

\(\displaystyle\int\limits_{0}^{2 \pi}(2 x+\cos x)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{2 \pi}2x\mathrm{\,d}x + \displaystyle\int\limits_{0}^{2 \pi}\cos x\mathrm{\,d}x=x^2 \Big|_0^{2\pi}+\sin x\Big|_0^{2\pi}=2\pi.\)

\(\displaystyle\int\limits_0^{\frac{\pi}{2}}(4 \sin x+3 \cos x) \mathrm{d} x = \left. -4 \cos x + 3 \sin x \right|_0^{\frac{\pi}{2}} = \left[-4 \cos\left(\displaystyle\frac{\pi}{2}\right) + 3 \sin\left(\displaystyle\frac{\pi}{2}\right) \right] - \left(-4 \cos 0 + 3 \sin 0 \right)= 7\).

\(\begin{aligned}\displaystyle\int\limits_0^{\tfrac{\pi}{4}}(\sin x+\cos x) \mathrm{\,d}x+\displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}(\sin x+\cos x) \mathrm{\,d}x &=\displaystyle\int\limits_0^{\tfrac{\pi}{2}}(\sin x+\cos x) \mathrm{\,d}x\\ &=\displaystyle\int\limits_0^{\tfrac{\pi}{2}} \sin x \mathrm{\,d}x+\displaystyle\int\limits_0^{\tfrac{\pi}{2}} \cos x \mathrm{\,d}x \\ &=(-\cos x)\Big|_0 ^{\tfrac{\pi}{2}}+(\sin x)\Big|_0 ^{\tfrac{\pi}{2}}=-(0-1)+(1-0)=2.\end{aligned}\)

\(\displaystyle\int\limits_{0}^{\frac{\pi}{2}} 2 \cos x \mathrm{~d} x=2\displaystyle\int\limits_{0}^{\frac{\pi}{2}} \cos x \mathrm{~d} x=2 \cdot 1=2\).

\(\displaystyle\int\limits_{\tfrac{\pi}{2}}^{-\tfrac{\pi}{2}} \cos x \mathrm{\,d} x=-\displaystyle\int\limits_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}} \cos x \mathrm{\,d} x=-(\sin x)\Big|_{-\tfrac{\pi}{2}} ^{\tfrac{\pi}{2}}=-\left[\sin \displaystyle\frac{\pi}{2}-\sin \left(-\displaystyle\frac{\pi}{2}\right)\right]=-2\).

\(\displaystyle\int\limits_0^{\tfrac{\pi}{2}}(3 \sin x-\cos x) \mathrm{\,d}x=3 \displaystyle\int\limits_0^{\tfrac{\pi}{2}} \sin x \mathrm{\,d}x-\displaystyle\int\limits_0^{\tfrac{\pi}{2}} \cos x \mathrm{\,d}x=(-3 \cos x)\bigg|_0 ^{\tfrac{\pi}{2}}-\sin x\bigg|_0 ^{\tfrac{\pi}{2}}\\=-3(0-1)-(1-0)=2\).

\(\displaystyle\int\limits_{0}^{\tfrac{\pi}{2}}(1+\sin x)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{\tfrac{\pi}{2}}\mathrm{\,d}x+\displaystyle\int\limits_{0}^{\tfrac{\pi}{2}}\sin x\mathrm{\,d}x=x\Big|_0^{\tfrac{\pi}{2}}-\cos x\Big|_0^{\tfrac{\pi}{2}}=\displaystyle\frac{\pi}{2}+1.\)

\(\displaystyle\int\limits_0^{\frac{\pi}{4}} \tan ^2 x \mathrm{~d} x = \int\limits_0^{\frac{\pi}{4}} \left(1 + \tan ^2 x \right) - 1\mathrm{~d} x = \int\limits_0^{\frac{\pi}{4}} \displaystyle\frac{1}{\cos^2 x} - \int\limits_0^{\frac{\pi}{4}} 1 \mathrm{~d} x=\left.\tan x \right|_0^{\frac{\pi}{4}} - \left. x \right|_0^{\frac{\pi}{4}}\).

\(= \left(\tan \displaystyle\frac{\pi}{4} - \tan 0\right) - (\displaystyle\frac{\pi}{4} - 0) = 1 - \displaystyle\frac{\pi}{4}\).

\(\displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \displaystyle\frac{-2}{3 \sin ^2 x} \mathrm{\,d} x=-\displaystyle\frac{2}{3} \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \displaystyle\frac{1}{\sin ^2 x} \mathrm{\,d} x=-\left.\displaystyle\frac{2}{3}(-\cot x)\right|_{\frac{\pi}{4}} ^{\frac{\pi}{2}}=\displaystyle\frac{2}{3}\left(\cot \displaystyle\frac{\pi}{2}-\cot \displaystyle\frac{\pi}{4}\right)=\displaystyle\frac{2}{3}(0-1)=-\displaystyle\frac{2}{3}\).

\(\displaystyle\int\limits_0^{\frac{\pi}{2}}(3 \sin x-2) \mathrm{\,d}x=-3\cos x\Big|_0^{\frac{\pi}{2}}-2x\Big|_0^{\frac{\pi}{2}}=3-\pi\).

\(\displaystyle\int\limits_0^{\frac{\pi}{2}} \displaystyle\frac{\sin ^2 x}{1+\cos x} \mathrm{\,d}x=\displaystyle\int\limits_0^{\frac{\pi}{2}} \displaystyle\frac{1-\cos ^2 x}{1+\cos x} \mathrm{\,d}x=\displaystyle\int\limits_0^{\frac{\pi}{2}} (1-\cos x) \mathrm{\,d}x=x\Big|_0^{\frac{\pi}{2}}-\sin x\Big|_0^{\frac{\pi}{2}}=\displaystyle\frac{\pi}{2}-1\).

\(\displaystyle\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\sin x| \mathrm{\,d} x=\displaystyle\int\limits_{-\frac{\pi}{2}}^{0}-\sin x \mathrm{\,d}x+\displaystyle\int\limits_{0}^{\frac{\pi}{2}}\sin x \mathrm{\,d}x=\cos x\Big|_{-\frac{\pi}{2}}^{0}-\cos x\Big|_{0}^{\frac{\pi}{2}}=2\).

\begin{eqnarray*}\displaystyle\int\limits_0^{2 \pi}|\sin x| \mathrm{d} x&=&\displaystyle\int\limits_0^\pi|\sin x| \mathrm{d} x+\displaystyle\int\limits_\pi^{2 \pi}|\sin x| \mathrm{d} x\\ &=&\displaystyle\int\limits_0^\pi \sin x \mathrm{~d} x-\displaystyle\int\limits_\pi^{2 \pi} \sin x \mathrm{~d} x\\ &=&-\cos x\bigg|_0 ^\pi+\cos x\bigg|_\pi ^{2 \pi}=4.\end{eqnarray*}

\(\displaystyle\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{4}}\displaystyle\frac{1}{\sin^2x}\mathrm{\,d}x=-\cot x\bigg|_{\frac{\pi}{6}}^{\frac{\pi}{4}}=-(1-\sqrt{3})=\sqrt{3}-1\).

\(\displaystyle\int\limits_{0}^{\frac{\pi}{4}}(1+\tan x)\cos x\mathrm{\,d}x=\displaystyle\int\limits_{0}^{\frac{\pi}{4}}(\cos x+\sin x)\mathrm{\,d}x=\left(\sin x-\cos x\right)\bigg|_{0}^{\frac{\pi}{4}}=1\).

\(\displaystyle\int\limits_{0}^{\frac{\pi}{4}} \displaystyle\frac{1}{\cos ^2x} \mathrm{\,d} x=\tan x\Big|_{0} ^{\frac{\pi}{4}}=\left(\tan \displaystyle\frac{\pi}{4}-\tan 0 \right)=1\).

\(\displaystyle\int\limits_0^{\frac{\pi}{4}}(\sin x+\cos x) \mathrm{d} x=\int\limits_0^{\frac{\pi}{4}} \sin x \mathrm{~d} x+\int\limits_0^{\frac{\pi}{4}} \cos x \mathrm{~d} x=-\cos x\bigg|_0 ^{\frac{\pi}{4}}+\sin x\bigg|_0 ^{\frac{\pi}{4}}\) \(=-\cos \displaystyle\frac{\pi}{4} - (-\cos 0) + \sin \displaystyle\frac{\pi}{4} - \sin 0 = -\displaystyle\frac{\sqrt{2}}{2} - (-1) + \displaystyle\frac{\sqrt{2}}{2} - 0 = 1.\)

\(\displaystyle\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{4}}\left(\displaystyle\frac{1}{\sin ^2 x}-\displaystyle\frac{1}{\cos ^2 x}\right) \mathrm{d} x = \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{4}} \displaystyle\frac{1}{\sin ^2 x} \mathrm{d} x- \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{4}}\displaystyle\frac{1}{\cos ^2 x} \mathrm{d} x = -\cot x \bigg|_{\frac{\pi}{6}}^{\frac{\pi}{4}} - \tan x\bigg|_{\frac{\pi}{6}}^{\frac{\pi}{4}}\) \(= -\cot \displaystyle\frac{\pi}{4}-\left(-\cot \displaystyle\frac{\pi}{6}\right) - \tan \displaystyle\frac{\pi}{4} -\left(-\tan \displaystyle\frac{\pi}{6}\right) = -1 + \sqrt{3} - 1 + \displaystyle\frac{1}{\sqrt{3}} = -2 + \displaystyle\frac{4\sqrt{3}}{3}\).

\(\begin{aligned}\displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{3}}\left(\displaystyle\frac{1}{\cos ^2 x}+\displaystyle\frac{\sqrt{3}}{\sin ^2 x}\right) \mathrm{\,d} x &=\displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{3}} \displaystyle\frac{1}{\cos ^2 x} \mathrm{\,d} x+\sqrt{3} \displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{3}} \displaystyle\frac{1}{\sin ^2 x} \mathrm{\,d} x\\ &=\tan x\Big|_{\tfrac{\pi}{4}} ^{\tfrac{\pi}{3}}+\left.\sqrt{3}(-\cot x)\right|_{\tfrac{\pi}{4}} ^{\tfrac{\pi}{3}}\\ &=\left(\tan \displaystyle\frac{\pi}{3}-\tan \displaystyle\frac{\pi}{4}\right)-\sqrt{3}\left(\cot \displaystyle\frac{\pi}{3}-\cot \displaystyle\frac{\pi}{4}\right)\\ &=(\sqrt{3}-1)-\sqrt{3}\left(\displaystyle\frac{1}{\sqrt{3}}-1\right)\\ &=2 \sqrt{3}-2.\end{aligned}\)

\(\begin{aligned}\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \displaystyle\frac{\cos2x}{\sin^{2}x\cos^{2}x}\mathrm{\,d}x\\ &=\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\displaystyle\frac{\cos2x}{(\sin x\cos x)^2}\mathrm{\,d}x\\ =\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\displaystyle\frac{\cos2x}{\left(\displaystyle\frac{1}{2}\sin 2x\right)^{2}}\mathrm{\,d}x=\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\displaystyle\frac{4\cos2x}{\sin^{2} 2x}\mathrm{\,d}x\\ &=\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\displaystyle\frac{\mathrm{\,d}(\sin^{2} 2x)}{\sin^{2}2x}=\left.{-\displaystyle\frac{1}{\tan 2x}} \right|_{\frac{\pi}{6}}^{\frac{\pi}{3}}=0.\end{aligned}\)

\(\displaystyle\int\limits_{\tfrac{\pi}{6}}^{\tfrac{\pi}{3}}\left(\displaystyle\frac{1}{\cos ^{2} x}-\displaystyle\frac{1}{\sin ^{2} x}\right) \mathrm{d} x=\displaystyle\int\limits_{\tfrac{\pi}{6}}^{\tfrac{\pi}{3}}\displaystyle\frac{1}{\cos ^{2} x} \mathrm{d} x-\displaystyle\int\limits_{\tfrac{\pi}{6}}^{\tfrac{\pi}{3}}\displaystyle\frac{1}{\sin ^{2} x} \mathrm{d} x=\tan x\Big|_{\tfrac{\pi}{6}}^{\tfrac{\pi}{3}} + \cot x\Big|_{\tfrac{\pi}{6}}^{\tfrac{\pi}{3}}=0\).

\(\displaystyle\int\limits_0^1\left(\mathrm{e}^x+1\right)\mathrm{\,d}x=\left(\mathrm{e}^x+x\right)\Biggr|_0^1=(\mathrm{e}+1)-(1+0)=\mathrm{e}\).

Ta có \(I=\displaystyle\int\limits_1^3 \mathrm{e}^x \mathrm{\,d}x=\mathrm{e}^x\Bigr|^{3}_{1}=\mathrm{e}^3-\mathrm{e}\).

Áp dụng công thức có \(I=\displaystyle \int_{0}^{1} 3^x \mathrm{\, d}x=\displaystyle\frac{3^x}{\ln3}\biggr|_0^1=\displaystyle\frac{2}{\ln3}\).

Ta có \(\displaystyle \int \limits_{1}^{2} \displaystyle\frac{1}{x^6} \mathrm{d} x = \int \limits_{1}^{2} x^{-6} \mathrm{d} x = \displaystyle\frac{x^{-5}}{-5} \bigg|_1^2 = \displaystyle\frac{31}{125}\).

Ta có \(I=\displaystyle\int\limits_0^12\mathrm{e}^x \mathrm{\,d}x=2\displaystyle\int_0^1\limits \mathrm{e}^x \mathrm{\,d}x=2(\mathrm{e}^x)\Big|_0^1=2\mathrm{e}-2\).

Ta có \( I=\displaystyle\int\limits_0^1 \mathrm{e}^x\mathrm{\,d}x=\mathrm{e}^x\big|_0^1=\mathrm{e}^1-\mathrm{e}^0=\mathrm{e}-1 \).

\(I=\displaystyle \int \limits_0^1 10^x \mathrm{d}x=\displaystyle\frac{10^x}{\ln 10} \Big|_0^1=\displaystyle\frac{9}{\ln 10}\).

Ta có \(\displaystyle \int\limits_{0}^{1} \mathrm{e}^x \mathrm{\,d}x = \mathrm{e}^x \big|_0^1 = \mathrm{e}^1 - \mathrm{e}^0 = \mathrm{e} - 1\).

\(I=\displaystyle\int\limits_{-1}^2f(x)\mathrm{\,d}x+\displaystyle\int\limits_{-1}^2 2x\mathrm{\,d}x=\left.F(x)\right|_{-1}^2+\left.x^2\right|_{-1}^2=F(2)-F(-1)+4-1=4-1+3=6.\)

Ta có \(I=\mathrm{e}^{2x} \bigg|_0^2 = \mathrm{e}^{4} - 1.\)

Ta có \(\displaystyle\int\limits_1^2 \mathrm{e}^{3x-1} \mathrm{\,d}x=\left(\displaystyle\frac{1}{3}\mathrm{e}^{3x-1}\right)\bigg|^2_1=\displaystyle\frac{1}{3}\left(\mathrm{e}^5-\mathrm{e}^2\right)\).

\(I=\displaystyle\int\limits_0^1 \mathrm{e}^{2x}\mathrm{\,d}x=\left. \displaystyle\frac{\mathrm{e}^{2x}}{2}\right| _0^1=\displaystyle\frac{\mathrm{e}^{2}-1}{2}.\)

Ta có \(\displaystyle\int\limits_0^2\mathrm{e}^{2x}\mathrm{\,d}x=\left.\displaystyle\frac{\mathrm{e}^{2x}}{2}\right|_0^2=\displaystyle\frac{1}{2}\mathrm{e}^{4}-\displaystyle\frac{1}{2}\).

Ta có \(I=\left(\displaystyle\frac{1}{4}\mathrm{e}^{4x}+x\right)\Bigg|_0^{\ln 2}=\left(\displaystyle\frac{1}{4}\mathrm{e}^{4\ln 2}+\ln 2\right)-\displaystyle\frac{1}{4}=\displaystyle\frac{15}{4}+\ln 2\).

Ta có \(\displaystyle\int\limits_0^1 \mathrm{e}^{-x}\mathrm{\,d}x=\left. -\mathrm{e}^{-x}\right|_0^1=\displaystyle\frac{\mathrm{e}-1}{\mathrm{e}}.\)

Ta có \(\displaystyle\int\limits_0^1{{\mathrm{e}^{3x + 1}}\mathrm{\,d}x}=\left.\displaystyle\frac{1}{3}\mathrm{e}^{3x+1}\right|_0^1=\displaystyle\frac{1}{3}(\mathrm{e}^4-\mathrm{e})\).

Ta có \(I=\displaystyle\int\limits_0^2 f(x)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x +\displaystyle\int\limits_{1}^{2} f(x)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{1} 1\mathrm{\,d}x+\displaystyle\int\limits_{0}^{1} x \mathrm{\,d}x =\displaystyle\frac{5}{2}\).

Ta có

\begin{align*}\displaystyle\int\limits_{0}^{2}f(x)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{1}3x^2\mathrm{\,d}x+\displaystyle\int\limits_{1}^{2}(4-x)\mathrm{\,d}x=\displaystyle\frac{7}{2}.\end{align*}

\(\displaystyle\int\limits_{0}^{\tfrac{\pi}{2}}\left(e^{x}-2 \cos x\right)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{\tfrac{\pi}{2}} e^{x}\mathrm{\,d}x-2 \displaystyle\int\limits_{0}^{\tfrac{\pi}{2}} \cos x\mathrm{\,d}x\) \(=\left.e^{x}\right|_{0} ^{\tfrac{\pi}{2}}-\left.2 \sin x\right|_{0} ^{\tfrac{\pi}{2}}=e^{\tfrac{\pi}{2}}-3\).

\(\displaystyle\int\limits_{1}^{4}\left(2^{x}-\displaystyle\frac{3}{x^{2}}\right) \mathrm{d} x=\displaystyle\int\limits_{1}^{4} 2^{x} \mathrm{~d} x-3 \displaystyle\int\limits_{1}^{4} x^{-2} \mathrm{~d} x=\left.\displaystyle\frac{2^{x}}{\ln 2}\right|_{1} ^{4}+\left.3 \cdot \displaystyle\frac{1}{x}\right|_{1} ^{4}=\displaystyle\frac{15}{\ln 2}-\displaystyle\frac{9}{4}\).

\(\displaystyle\int\limits_{1}^{2}\left(3^{x}-\displaystyle\frac{3}{x}\right)\mathrm{\,d}x=\displaystyle\int\limits_{1}^{2}3^{x}\mathrm{\,d}x-\displaystyle\int\limits_{1}^{2}\displaystyle\frac{3}{x}\mathrm{\,d}x=\displaystyle\frac{6}{\ln3}-3\ln2\).

\(\displaystyle\int\limits_{0}^{1}\left(e^{2 x}+3 x^{2}\right)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{1}e^{2x}\mathrm{\,d}x+\displaystyle\int\limits_{0}^{1}3x^2\mathrm{\,d}x=\displaystyle\frac{1}{2}e^{2x}\Big|_0^1+x^3\Big|_0^1=\displaystyle\frac{1}{2}\left(e^2-1\right)+(1-0)=\displaystyle\frac{1}{2}e^2+\displaystyle\frac{1}{2}\).

Theo đề bài ta có \(f'(x)=(x-1)^2\), suy ra

\[f(x)= \int (x-1)^2\mathrm{\,d}x=\int (x-1)^2\mathrm{\,d}(x-1)=\displaystyle\frac{(x-1)^3}{3}+C\]

Vì \(M\) trùng với gốc tọa độ khi nó nằm trên trục tung nên

\( f(0)=0\Leftrightarrow C=\displaystyle\frac{1}{3}.\)

Vậy \(f(x)=\displaystyle\frac{(x-1)^3}{3}+\displaystyle\frac{1}{3}\).

\(\begin{aligned}\displaystyle\int_{-5}^{0}\left(3^{x+1}-2e^x\right) \mathrm{\,d}x\\&=\displaystyle\int_{-5}^{0}\left(3\cdot3^{x}-2e^x\right) \mathrm{\,d}x=\left. {3\cdot\displaystyle\frac{3^{x}}{\ln3}-2e^{x}}\right|_{-5}^{0}=\displaystyle\frac{3}{\ln3}-2-\left(\displaystyle\frac{1}{81\cdot\ln3}-\displaystyle\frac{2}{e^{5}} \right)\\&=\left(3-\displaystyle\frac{1}{81}\right)\cdot\displaystyle\frac{1}{\ln3}-2+\displaystyle\frac{2}{e^{5}}=\displaystyle\frac{242}{81\cdot\ln3}+\displaystyle\frac{2}{e^{5}}-2.\end{aligned}\)

\(\displaystyle\int_{1}^{2}2^x\cdot3^{x-1}\mathrm{\,d}x=\displaystyle\int_{1}^{2}\frac{6^x}{3}\mathrm{\,d}x=\left. {\displaystyle\frac{1}{3}\cdot\displaystyle\frac{6^x}{\ln 6}} \right|_{1}^{2}=\displaystyle\frac{12}{\ln 6}-\displaystyle\frac{2}{\ln 6}=\displaystyle\frac{10}{\ln 6}\).

\(\displaystyle\int\limits_0^1 \mathrm{e}^t \mathrm{~d} t=\mathrm{e}^t\bigg|_0 ^1=\mathrm{e}^1-\mathrm{e}^0=e-1\).

Ta có \(F(\mathrm{e})-F(1)=\displaystyle\int\limits_1^\mathrm{e} f(x) \mathrm{d} x=\displaystyle\int\limits_1^\mathrm{e} \frac{1}{x} \mathrm{~d} x=\ln x\bigg|_1 ^\mathrm{e}=1\).

Vậy \(F(\mathrm{e})=F(1)+1=1+1=2\).

Ta có \(\displaystyle\int\limits_0^2 \frac{\sqrt{\mathrm{e}^x}}{2} \mathrm{~d} x=\frac{1}{2} \displaystyle\int\limits_0^2 \sqrt{\mathrm{e}^x} \mathrm{~d} x=\frac{1}{2} \cdot 2(\mathrm{e}-1)=\mathrm{e}-1\).

\(\displaystyle\int\limits_0^2 3^x \mathrm{\,d} x=\left. \displaystyle\frac{3^x}{\ln 3}\right|_0 ^2=\displaystyle\frac{1}{\ln 3}(9-1)=\displaystyle\frac{8}{\ln 3}\).

\(\displaystyle\int\limits_1^e \displaystyle\frac{1}{t} \mathrm{\,d} t=\ln t\big|_1^e=\ln e-\ln 1=1\).

\(\displaystyle\int\limits_0^2 2^{x+3} \mathrm{\,d} x=\displaystyle\int\limits_0^2 2^x \cdot 2^3 \mathrm{\,d} x=8 \displaystyle\int\limits_0^2 2^x \mathrm{\,d} x=\left.8 \cdot \displaystyle\frac{2^x}{\ln 2}\right|_0 ^2=\displaystyle\frac{8}{\ln 2}(4-1)=\displaystyle\frac{24}{\ln 2}\).

Đặt \(u=3x+2\) ta có \(\mathrm{d}u\,=3\mathrm{d}x\,\) và khi \(x=-1\Rightarrow u=-1\). \(x=0\Rightarrow u=2\).

Khi đó \(\displaystyle\int\limits_{-1}^02^{3x+2}\mathrm{d}x\,=\displaystyle\frac13\displaystyle\int\limits_{-1}^22^u\mathrm{d}u\,=\displaystyle\frac13\cdot\displaystyle\frac{2^u}{\ln 2}\Big|_{-1}^2=\displaystyle\frac1{3\ln 2}(2^2-2^{-1})=\displaystyle\frac{7}{6\ln2}\).

\(\displaystyle\int\limits_0^22^x\cdot3^{x+1}\mathrm{d}x\,=\displaystyle\int\limits_0^23\cdot6^x\mathrm{d}x\,=\displaystyle\frac2{\ln6}6^x\Big|_0^2=\displaystyle\frac2{\ln6}(6^2-6^0)=\displaystyle\frac{70}{\ln6}\).

\(\displaystyle\int\limits_0^1\displaystyle\frac{7^x}{11^x}\mathrm{d}x\,=\displaystyle\int\limits_0^1\left(\displaystyle\frac7{11}\right)^x\mathrm{d}x\,=\displaystyle\frac1{\ln\displaystyle\frac7{11}}\left(\displaystyle\frac7{11}\right)^x\Big|_0^1=\displaystyle\frac1{\ln7-\ln11}\left(\displaystyle\frac7{11}-1\right)=\displaystyle\frac4{11(\ln11-\ln7)}\).

\(\displaystyle\int\limits_{-1}^0 \mathrm{e}^{-x} \mathrm{~d} x = \left. -\mathrm{e}^{-x} \right|_{-1}^0 = -\mathrm{e}^0 - (-\mathrm{e}^1) = \mathrm{e} - 1\).

\(\displaystyle\int\limits_{-2}^{-1} \mathrm{e}^{x+2} \mathrm{~d} x = \mathrm{e}^2 \int\limits_{-2}^{-1} \mathrm{e}^{x} \mathrm{~d} x = \mathrm{e}^2 \left.(\mathrm{e}^x)\right|_{-2}^{-1} = \mathrm{e}^2(\mathrm{e}^{-1}-\mathrm{e}^{-2})=\mathrm{e}-1\).

\(\displaystyle\int\limits_0^1\left(3 \cdot 4^x-5 \mathrm{e}^{-x}\right) \mathrm{d} x = \left. 3\cdot\displaystyle\frac{4^x}{\ln 4}+ 5 \mathrm{e}^{-x} \right|_0^1 = \left(\displaystyle\frac{12}{\ln 4} + \displaystyle\frac{5}{e}\right) - \left(\displaystyle\frac{3}{\ln 4} + 5 \right) = \displaystyle\frac{5}{e}-5 + \displaystyle\frac{9}{\ln 4}\).

\(\displaystyle\int\limits_0^1 \mathrm{e}^x \mathrm{~d} x=\left.\mathrm{e}^x\right|_0 ^1=\mathrm{e}^1-\mathrm{e}^0=\mathrm{e}-1\).

}

\(\displaystyle\int\limits_0^1 \mathrm{e}^x \mathrm{~d} x=\left.\mathrm{e}^x\right|_0 ^1=\mathrm{e}^1-\mathrm{e}^0=\mathrm{e}-1\).

\item \(\displaystyle\int\limits_0^1 2^x \mathrm{~d} x=\left.\frac{2^x}{\ln 2}\right|_0 ^1=\frac{2^1}{\ln 2}-\frac{2^0}{\ln 2}=\frac{1}{\ln 2}\).

}

\(\begin{aligned}\displaystyle \int\limits_0^1\left(3 \cdot 2^x-\mathrm{e}^x\right) \mathrm{d} x & =\int\limits_0^1 3 \cdot 2^x \mathrm{~d} x-\int\limits_0^1 \mathrm{e}^x \mathrm{~d} x=3 \int\limits_0^1 2^x \mathrm{~d} x-\int\limits_0^1 \mathrm{e}^x \mathrm{~d} x\\& =\frac{3}{\ln 2}-(e-1)=\frac{3}{\ln 2}-e+1.\end{aligned}\)

}

Tích phân cần tính là diện tích của hình thang vuông \(O A B C\), có đáy nhỏ \(O C=1\), đáy lớn \(A B=2\) và đường cao \(O A=1\) (xem hình).

Do đó

\(\displaystyle\int\limits_{0}^{1}(x+1) \mathrm{d} x=S_{O A B C}=\displaystyle\frac{1}{2}(O C+A B) \cdot O A=\displaystyle\frac{1}{2}(1+2) \cdot 1=\displaystyle\frac{3}{2}.\)

Ta có \(y=\sqrt{1-x^{2}}\) là phương trình nửa phía trên trục hoành của đường tròn tâm tại gốc toạ độ \(O\) và bán kính 1. Do đó, tích phân cần tính là diện tích nửa phía trên trục hoành của hình tròn như hình vẽ.

Vậy \(\displaystyle\int\limits_{-1}^{1} \sqrt{1-x^{2}}\mathrm{\,d}x=\displaystyle\frac{\pi}{2}\).

Ta có \(y=\sqrt{9-x^{2}}\) là phương trình nửa phía trên trục hoành của đường tròn tâm tại gốc toạ độ \(O\) và bán kính \(3\). Do đó, tích phân cần tính là diện tích nửa phía trên trục hoành của hình tròn như hình vẽ.

Do đó

\(\displaystyle\int\limits_{-2}^{2} \sqrt{4-x^{2}}\mathrm{\,d}x=\displaystyle\frac{1}{2}\cdot \pi\cdot 3^2=\displaystyle\frac{9\pi}{2}.\)

Tích phân cần tính là diện tích hình thang vuông $ABCD$ (hình vẽ), có đáy nhỏ $AD=3$, đáy lớn $BC=5$, chiều cao $AB=1$.

Do đó

\(\displaystyle\int\limits_{1}^{3}(2 x+1)\mathrm{\,d}x=S_{ABCD}=\dfrac{1}{2}\cdot (AD+BC)\cdot AB=4.\)

Ta có \(Q(t)=\displaystyle \int Q'(t)\mathrm{\,d}t=\int \left(3t^2-6t+5\right)\mathrm{\,d}t = t^3-3t^2+5t+C\).

Vì \(Q(1)=4\Rightarrow 1-3+5+C=4\Leftrightarrow C=1\Rightarrow Q(t)=t^3-3t^2+5t+1\).

Suy ra điện lượng truyền trong dây dẫn khi \(t=3\) là \(Q(3)=3^3-3\cdot 3^2+5\cdot 3+1=16\) (C).

a) Ta có \(f'(x) = \left( \ln y(x) \right)' = \displaystyle\frac{y'(x)}{y(x)} = -7 \cdot 10^{-4}\). Suy ra \(f(x) = \displaystyle\int\limits -7 \cdot 10^{-4} \mathrm{~d}x = -7 \cdot 10^{-4} x + C\).

Tại \(x = 0\), ta có \(f(0) = \ln y(0) = \ln(0{,}05)\), suy ra \(C = \ln(0{,}05)\).

Vậy \(f(x) = -7 \cdot 10^{-4} x + \ln(0{,}05) \).

b) Ta có \(f(x) = \ln y(x)\) với \(x \geq 0\), suy ra \(y(x) = \mathrm{e}^{f(x)} = \mathrm{e}^{-7 \cdot 10^{-4} x + \ln(0{,}05)}= 0{,}05 \mathrm{e}^{-7 \cdot 10^{-4}x}\).

Nồng độ trung bình của chất \(A\) từ thời điểm 15 giây đến thời điểm 30 giây là:

\[\displaystyle\frac{1}{30-15} \displaystyle\int\limits_{15}^{30} 0{,}05 \mathrm{e}^{-7 \cdot 10^{-4}x } \mathrm{~d}x = \displaystyle\frac{0{,}05}{15} \left. \displaystyle\frac{ \mathrm{e}^{-7 \cdot 10^{-4}x }}{-7 \cdot 10^{-4}} \right|_{15}^{30} \approx 0{,}04922.\]

Quãng đường vật di chuyển trong khoảng thời gian từ thời điểm \(t=0\) (s) đền thời điểm \(t = \displaystyle\frac{3\pi}{4}\) (s) là:

\[\displaystyle\int\limits_0^{\frac{3\pi}{4}} 2 - \sin t \mathrm{~d}t = (2t+\cos t)\bigg|_0^{\frac{3\pi}{4}} = \left(\displaystyle\frac{6\pi}{4}+\cos\displaystyle\frac{3\pi}{4}\right)-(0+\cos 0)=\displaystyle\frac{3\pi}{2} - \displaystyle\frac{\sqrt{2}}{2}-1\, (m).\]

a) Quãng đường mà vật di chuyển được trong 1 giây đầu tiên bằng diện tích tam giác \(OAD\):

\[S_{OAD} = \displaystyle\frac{1}{2}\cdot 1 \cdot 2 = 1 (m).\]

b) Quãng đường mà vật di chuyển được trong 2 giây đầu tiên bằng diện tích hình thang \(OABC\):

\[S_{OABC} = \displaystyle\frac{1}{2}\cdot 2 \cdot (2+1) = 3 (m).\]

a) Công thức xác định hàm số \(h(t)\) là

\begin{align*}h(t)&=\displaystyle\int v(t)\mathrm{d}t\,=\displaystyle\int(-0{,}12t^2+1{,}2t)\mathrm{d}t\,\\&=-\displaystyle\frac{0{,}12}3t^3+\displaystyle\frac{1{,}2}2t^2+C=-0{,}04t^3+0{,}6t^2+C.\end{align*}

Vì \(h(0)=520\) nên \(C=520\).

Vậy \(h(t)=-0{,}04t^3+0{,}6t^2+520\).

b) Ta có \(h'(t)=v(t)=-0{,}12t^2+1{,}2t=0\Leftrightarrow\hoac{&t=0\\&t=10.}\)

Lập bảng biến thiên, ta được

Vậy độ cao của khinh khí cầu khi bay là \(540\) mét.

c) Khi khí cầu trở lại độ cao khi xuất phát ứng với \(t\) là nghiệm của phương trình

\(h(t)=h(0)\Leftrightarrow -0{,}04t^3+0{,}6t^2+520=520\Leftrightarrow t=0\) hoặc \(t=15.\)

Vậy \(15\) phút sau khi xuất phát thì khinh khí cầu trở lại độ cao khi xuất phát.

a) Khi có \(360\) công nhân được sử dụng ta có

\begin{align*}m(t)=360&\Leftrightarrow 500+50\sqrt t-10t=360\\ &\Leftrightarrow 10t-50\sqrt t-140=0\\&\Leftrightarrow\left[\begin{aligned}&\sqrt t=7\\&\sqrt t=-2\text{ (loại)}\end{aligned}\right.\Leftrightarrow t=49.\end{align*}

Vậy sau \(t\) ngày thì có \(360\) công nhân được sử dụng.

b) Ta có \(m'(t)=50\cdot\displaystyle\frac1{2\sqrt t}-10=0\Leftrightarrow \sqrt t=\displaystyle\frac{25}{10}=2{,}5\Leftrightarrow t=2{,}5^2=6{,}25\).

Ta có \(m(0)=500\); \(m(2)=560\); \(m(3)=560\); \(m(10)=0\).

Vậy số công nhân được sử dụng nhiều nhất vào ngày thứ \(2\) và ngày thứ \(3\).

c) Vì \(M'(t)=m(t)\) nên tổng số ngày công để hoành thành công trình đó là

\(\int\limits_0^{100}M'(t)\mathrm{\,d}t=\int\limits_0^{100}(500+50\sqrt t-10t)\mathrm{\,d}t\approx33~333\) (ngày).

Tốc độ lây lan của virus cúm ở ngày thứ \(t\) là \(P'(t)\) nên ta có

\begin{align*}P'(t)=k(1~000-P(t))&\Leftrightarrow P'(t)+k\cdot P(t)=k\cdot1~000\\ &\Rightarrow (\mathrm{e}^{kt}P(t))'=1~000k\mathrm{e}^{kt}\\ &\Rightarrow\mathrm{e}^{kt}P(t)=\int1~000k\mathrm{e}^{kt}\mathrm{\,d}t=1~000\mathrm{e}^{kt}+C\\ &\Rightarrow P(t)=1~000+C\cdot\mathrm{e}^{-kt}.\end{align*}

Vì \(P(4)=52\) nên \(1~000+C\cdot\mathrm{e}^{-4k}=52\Leftrightarrow C=-948\mathrm{e}^{4k}\).

Vậy, \(P(10)=1~000-948\mathrm{e}^{4k-k\cdot10}=1~000-948\mathrm{e}^{-6k}\).

a) Từ dữ liệu đề bài, ta có hệ

\(\begin{cases}d=0\\10^3a+10^2b+10c+d=5\\20^3a+20^2b+20c+d=21\\30^3a+30^2b+30c+d=40\end{cases}\Leftrightarrow\begin{cases}a=-\displaystyle\frac1{750}\\b=\displaystyle\frac{19}{200}\\c=-\displaystyle\frac{19}{60}\\d=0.\end{cases}\)

Từ đó ta có \(f(x)=-\displaystyle\frac1{750}x^3+\displaystyle\frac{19}{200}x^2-\displaystyle\frac{19}{60}x\).

b) Quãng đường mà xe ô tô đó đã đi được tính đến giây thứ \(60\) là

\(S=\int\limits_0^60f(t)\mathrm{\,d}t=1~950\) (mét).

a) \(S(3)=\displaystyle\frac{(2+4)\cdot 2}{2}=6\).

b) \(S(x)=\displaystyle\frac{(2+x+1)\cdot (x-1)}{2}=\displaystyle\frac{(x+3)(x-1)}{2}=\displaystyle\frac{x^2+2x-3}{2}\) (Với mỗi \(x\geq1\)).

c) Ta có \(S'(x)=\left(\displaystyle\frac{x^2+2x-3}{2}\right)'=\displaystyle\frac{2x+2}{2}=x+1=f(x)\).

Suy ra \(S(x)\) là một nguyên hàm của \(f(x)\) trên \([1;+\infty)\).

d) Ta có \(F(x)=\displaystyle \int f(x)\mathrm{\,d}x=\int (x+1)\mathrm{\,d}x=\displaystyle\frac{x^2}{2}+x+C\), với \(C\) là hằng số.

Ta có \(F(3)-F(1)=\left(\displaystyle\frac{3^2}{2}+3+C\right)-\left(\displaystyle\frac{1^2}{2}+1-C\right)=6=S(3)\) (đpcm).

Vận tốc xe tại thời điểm \(t\) giây là \(v(t)=\displaystyle\int a(t)\mathrm{\,d}t=\int 2\mathrm{\,d}t=2t+C\).

Ta có \(v(0)=10\Leftrightarrow C=10\).

Do đó \(v(t)=2t+10\) (m/s).

Quãng đường xe đi được sau \(t\) giây là \(s(t)=\displaystyle\int v(t)\mathrm{\,d}t=\int (2t+10)\mathrm{\,d}t=t^2+10t+C\) (m).

Lúc bắt đầu tăng tốc (tại thời điểm \(t=0\)) thì \(s=0\) nên ta có \(0^2+10\cdot 0+C=0\Rightarrow C=0\).

Suy ra \(s(t)=t^2+10t\).

Từ đó suy ta \(s(3)=3^2+10\cdot 3=39\) (mét).

Ta có:

\(\begin{aligned}C(100)-C(0) &=\displaystyle\int\limits_0^{100} C^{\prime}(x) \mathrm{\,d} x=\displaystyle\int\limits_0^{100}\left(5-0{,}06 x+0{,}00072 x^2\right) \mathrm{\,d} x \\&=5 \displaystyle\int\limits_0^{100} \mathrm{\,d} x-0{,}06 \displaystyle\int\limits_0^{100} x \mathrm{\,d} x+0{,}00072 \displaystyle\int\limits_0^{100} x^2 \mathrm{\,d} x \\&=\left.5 x\right|_0 ^{100}-0{,}03 x^2\left.\right|_0 ^{100}+0{,}00024 x^3\left.\right|_0 ^{100}=440.\end{aligned}\)

Suy ra \(C(100)=C(0)+440=30+440=470\) (triệu đồng).

Vậy khi nhà máy sản xuất 100 tấn sản phẩm A trong tháng thì tổng chi phí là \(470\) triệu đồng.

Ta có \(T(6)=150\) (khí bên trong ống được duy trì ở \(150^{\circ} C\)).

\(\begin{aligned}T(8)-T(6) &=\displaystyle\int\limits_6^{8} T^{\prime}(x) \mathrm{\,d}x=\displaystyle\int\limits_6^{8}-\displaystyle\frac{30}{x} \mathrm{\,d} x \\&=-30 \displaystyle\int\limits_6^{8}\displaystyle\frac{1}{x} \mathrm{\,d} x \\&=\left.-30\ln x\right|_6 ^{8}\approx -8{,}63.\end{aligned}\)

Suy ra \(T(8)=T(6)-8{,}63\approx 141{,}37\) (\(^{\circ} C\)).

Vậy nhiệt độ mặt ngoài của ống là \(141{,}37^{\circ} C\).

Quãng đường chuyển động của thang máy là

\(\displaystyle\int\limits_0^{24} v(t)\mathrm{\,dt} =\displaystyle\int\limits_0^2 t \mathrm{\,dt}+\displaystyle\int\limits_2^{20}2 \mathrm{\,dt}+\displaystyle\int\limits_{20}^{24}(12-0{,}5t)\mathrm{\,dt}=42\,(\mathrm{m}).\)

Tốc độ trung bình của thang máy là

\(\displaystyle\frac{1}{24-0} \displaystyle\int\limits_0^{24} v(t) \mathrm{\,d} t=\displaystyle\frac{42}{24}=1{,}75 \,(\mathrm{m/s}).\)

Quãng đường vật đi được trong khoảng thời gian trên là

\(s=\displaystyle\int\limits_{0}^{5}v(t)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{5}(3t+4)\mathrm{\,d}x=\left(\displaystyle\frac{3t^2}{2}+4t\right)\bigg|_{0}^{5}=\displaystyle\frac{75}{2}+20=\displaystyle\frac{115}{2}\).

Vì \(a(t)=v'(t)\) nên

\(v(t)= \displaystyle \int a(t) \mathrm{d}t=\displaystyle \int 3 \mathrm{d}t=3t+C\)

Ta có: \(v_0=1\) nên \(3.0+C=1\) hay \(C=1\). Vậy \(v(t)=3t+1 \text{ m/s}\)

Vậy tốc độ của chất điểm sau \(10\) giây kể từ khi bắt đầu tăng tốc là \(v(10)=3.10+1=31 \text{ m/s}\)

Ta có \(P(t)=\displaystyle \int P'(t) \mathrm{d}t=\displaystyle \int 20.(1,106)^t \mathrm{d}t=20.\displaystyle\frac{1,106^t}{\ln{1,106}}+C\)

Vì dân số thành phố đầu năm \(2015\text{ }(t=0)\) là \(1008\) nghìn người nên

\(20.\displaystyle\frac{1,106^0}{\ln{1,106}}+C=1008\)

hay \(C \approx 810\)

Vậy \(P(t)=20.\displaystyle\frac{1,106^t}{\ln{1,106}}+810\)

a) Dân số của thành phố ở thời điểm đầu năm \(2020 \text{ } (t=5)\) là

\(P(5)=20.\displaystyle\frac{1,106^5}{\ln{1,106}}+810 \approx 1139\) (nghìn người).

b) Tốc độ tăng dân số trung bình hằng năm của thành phố trong giai đoạn từ đầu năm \(2015\) đến đầu năm \(2020\) là

\(\displaystyle\frac{1139-1008}{5}=26,2\) (nghìn người).

Vì \(v(t)=s'(t)\) nên \(s(t)=\displaystyle \int v(t)\mathrm{d}t=\displaystyle \int 10t\mathrm{d}t=5t^2+C\)

Ta có \(s(0)=0 \text{ (m) nên } 5.0^2+C=0\) hay \(C=0\).

a) Vậy quãng đường vật di chuyển được sau thời gian \(t\) giây là \(s(t)=5t^2\)

b) Thời gian vật chạm đất là \(s(t)=100 \text{ hay } 5t^2=100\)

\(\Leftrightarrow \left [\begin{array}{l}t=2\sqrt{5} \text{ (nhận)}\\ t=-2\sqrt{5} \text{ (loại)}\end{array} \right.\)

Công tối thiểu để phóng một vệ tinh nặng \(m=1000\) kg từ mặt đất lên độ cao \(35\,780\) km so với mặt đất là

\(W=\displaystyle\int_{6370000}^{42150000}F(h)\mathrm{\,d}h=\displaystyle\int_{6370000}^{42150000}6{,}67\cdot10^{-11}\cdot\displaystyle\frac{6\cdot10^{24}\cdot1000}{42150000^{2}}\mathrm{\,d}h=8\,059\,762\,837\) (J).

Thời gian ô tô chuyển động cho đến khi dừng lại là \(v(t)=0\Leftrightarrow-5t+20=0\Leftrightarrow t=4\) (s).

Từ lúc đạp phanh đến khi dừng hẳn, ô tô còn di chuyển thêm một quãng đường dài là

\(\displaystyle\int_{0}^{4}v\left(t\right)\mathrm{\,d}t=\displaystyle\int_{0}^{4}\left(-5t+20\right)\mathrm{\,d}t=40\) (m).

Lượng nước chảy ra khỏi bồn trong \(10\) phút đầu từ khi bồn bị rỉ nước là

\(\displaystyle\int_{0}^{10}V^{\prime}(t)\mathrm{\,d}t=\displaystyle\int_{0}^{10}\left(200-4t\right) \mathrm{\,d}t=1800\) (lít).

Tổng chi phí sản xuất tăng lên nếu sản phẩm sản xuất ra tăng từ \(40\) sản phẩm lên \(50\) sản phẩm là

\(\displaystyle \int_{40}^{50}C^{\prime}(x) \mathrm{\,d}x=\displaystyle \int_{40}^{50}\left( 2x+80\right)\mathrm{\,d}x=1700\) (USD).

Cân nặng của một bé gái \(5\) tháng tuổi là

\(\displaystyle\int_{0}^{5}f^{\prime}(t)\mathrm{\,d}t=\displaystyle\int_{0}^{5}\left( 0{,}00093t^{2}-0{,}04792t+0{,}76806\right) \mathrm{\,d}t+3{,}3=6{,}58005\) (kg).

a) Độ dịch chuyển của vật trong khoảng thời gian \(1 \leq t \leq 4\) là \(\displaystyle\int\limits_{1}^{4} (t^{2}-t-6) \mathrm{d} t=-\displaystyle\frac{9}{2}.\)

b) Tổng quãng đường vật đi được trong khoảng thời gian \(1 \leq t \leq 4\) là \(\displaystyle\int\limits_{1}^{4} |t^{2}-t-6|\mathrm{d} t=\displaystyle\frac{61}{6}.\)

a) Ta có \(\displaystyle\int\limits_{100}^{101} P'(x)\mathrm{\,d}x \approx 12{,}5\) nên khi doanh số tăng từ 100 lên 101 đơn vị sản phẩm thì lợi nhuận tăng khoảng \(12,5\) triệu đồng.

b) Ta có \(\displaystyle\int\limits_{100}^{110} P'(x)\mathrm{\,d}x\approx 121{,}5\) nên khi doanh số tăng từ 100 lên 110 đơn vị sản phẩm thì lợi nhuận tăng khoảng \(121{,}5\) triệu đồng.

Vận tốc trung bình của động mạch trong khoảng \(0\leq r\leq R\) là

\(v_{tb}=\displaystyle\frac{1}{R}\int\limits_{0}^R k(R^2-r^2)\mathrm{\,d}r=\displaystyle\frac{k}{R}\left(R^2r-\displaystyle\frac{1}{3}r^3\right)\Big|_0^R=\displaystyle\frac{2}{3}kR^2.\)

Ta có \(v\) lớn nhất khi \(r=0\), tức là \(v_{\max}=kR^2\), đó đó \(v_{tb}=\displaystyle\frac{2}{3}v_{\max}\).

Ta có gia tốc của viên đạn là \(a(t)= -9{,}8t\).

Vận tốc của viên đạn là \(v(t) = \displaystyle \int a(t)dt = \displaystyle \int \left(-9{,}8t\right)dt = -\displaystyle\frac{9{,}8}{2}t^2 +C\).

Vận tốc thời điểm ban đầu của viên đạn là \(30\) (m/s) ta có \(v(0) = 30 \Leftrightarrow C = 30\).

Suy ra \(v(t) = -\displaystyle\frac{9{,}8}{2}t^2 + 30\).

Vậy vận tốc của viên đạn ở thời điểm \(2\)(s) là \(v(2)= 10{,}4\) (m/s).

Ta có khoảng cách mà con cá bơi được là

\[s(t)=\displaystyle \int v(t)dt=\displaystyle\int \left(\displaystyle\frac{-2t}{5}+4\right)dt= -\displaystyle\frac{t^2}{5} + 4t +C.\]

Thời điểm \(t=0\) cá bắt đầu bơi vào sông, ta có \(s(0)=0 \Leftrightarrow C=0\).

Ta có khoảng cách cá bơi là \(s(t) = -\displaystyle\frac{t^2}{5} + 4t \).

Vậy khoảng cách mà cá bơi được xa nhất là \(20\) (km) với \(t=10\) (s).