Tính tích phân :
\(\int^1_0x^3e^{x^2}dx\)
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PT
1
4 tháng 4 2016
\(=\frac{1}{2}\int\limits^1_0\ln\left(1+x^2\right)d\left(1+x^2\right)=\frac{1}{2}\left[\left(1+x^2\right)\ln\left(1+x^2\right)\right]|^1_0-\int\limits^1_0d\left(1+x^2\right)\)
\(=\frac{1}{2}\left[2\ln2-\left(1+x^2\right)|^1_0\right]=\frac{\left(2\ln2-1\right)}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 11 2021
Xét \(I=\int\limits^1_0x^2f\left(x\right)dx\)
Đặt \(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=x^2dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=\dfrac{1}{3}x^3\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{1}{3}x^3.f\left(x\right)|^1_0-\dfrac{1}{3}\int\limits^1_0x^3.f'\left(x\right)dx=-\dfrac{1}{3}\int\limits^1_0x^3f'\left(x\right)dx\)
\(\Rightarrow\int\limits^1_0x^3f'\left(x\right)dx=-1\)
Lại có: \(\int\limits^1_0x^6.dx=\dfrac{1}{7}\)
\(\Rightarrow\int\limits^1_0\left[f'\left(x\right)\right]^2dx+14\int\limits^1_0x^3.f'\left(x\right)dx+49.\int\limits^1_0x^6dx=0\)
\(\Rightarrow\int\limits^1_0\left[f'\left(x\right)+7x^3\right]^2dx=0\)
\(\Rightarrow f'\left(x\right)+7x^3=0\)
\(\Rightarrow f'\left(x\right)=-7x^3\)
\(\Rightarrow f\left(x\right)=\int-7x^3dx=-\dfrac{7}{4}x^4+C\)
\(f\left(1\right)=0\Rightarrow C=\dfrac{7}{4}\)
\(\Rightarrow I=\int\limits^1_0\left(-\dfrac{7}{4}x^4+\dfrac{7}{4}\right)dx=...\)
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 3 2023
Đặt \(2x+2=u\Rightarrow2xdx=du\Rightarrow dx=\dfrac{1}{2}du\)
\(\left\{{}\begin{matrix}x=0\Rightarrow u=2\\x=2\Rightarrow u=6\end{matrix}\right.\)
\(\Rightarrow I=\int\limits^6_2f\left(u\right).\dfrac{1}{2}du=\dfrac{1}{2}\int\limits^6_2f\left(u\right)du=\dfrac{1}{2}\int\limits^6_2f\left(x\right)dx=\dfrac{1}{2}.6=3\)
H
5 tháng 6 2023
\(\int\limits^1_0\dfrac{dx}{\left(x^2+3x+2\right)^2}=\int\limits^1_0\left(\dfrac{1}{x+1}-\dfrac{1}{x+1}\right)^2dx\)
\(=\int\limits^1_0\dfrac{dx}{\left(x+1\right)^2}+\int\limits^1_0\dfrac{dx}{\left(x+2\right)^2}-2\int\limits^1_0\dfrac{dx}{\left(x+1\right)\left(x+2\right)}\)
\(=-\dfrac{1}{x+1}\left|^1_0-\dfrac{1}{x+2}\right|^1_0-2\int\limits^1_0\dfrac{dx}{x+1}+2\int\limits^1_0\dfrac{dx}{x+2}\)
\(=\dfrac{2}{3}-4ln2+2ln3\)
\(\int\limits^1_0x^3e^{x^2}dx=\int\limits^1_0x^3e^{x^2}.xdx\)
Đặt \(t=x^2\Rightarrow\begin{cases}dt=2xdx;x=0\rightarrow t=0,x=1\rightarrow t=1\\f\left(x\right)dx=te^tdt\end{cases}\)
Do đó : \(I=\int\limits^1_0te^1dt=\frac{1}{2}\int\limits^1_0t.d\left(e^t\right)=\frac{1}{2}\left(t.e^t-e^t\right)|^1_0=\frac{1}{2}\)