Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
\(F'\left(x\right)=e^x+\left(x-1\right)e^x=xe^x=\frac{x}{e^x}.e^{2x}\Rightarrow f\left(x\right)=\frac{x}{e^x}\)
Xét \(I=\int f'\left(x\right)e^{2x}dx\)
Đặt \(\left\{{}\begin{matrix}u=e^{2x}\\v=f'\left(x\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2e^{2x}dx\\v=f\left(x\right)\end{matrix}\right.\)
\(\Rightarrow I=f\left(x\right).e^{2x}+2\int f\left(x\right).e^{2x}dx=x.e^x+2\left(x-1\right)e^x+C=\left(3x-2\right)e^x+C\)
2.
Xét \(J=\int\limits^1_0xf\left(6x\right)dx\)
Đặt \(6x=t\Rightarrow dx=\frac{1}{6}dt\Rightarrow J=\frac{1}{36}\int\limits^6_0t.f\left(t\right)dt=\frac{1}{36}\int\limits^6_0x.f\left(x\right)dx=1\)
\(\Rightarrow I=\int\limits^6_0x.f\left(x\right)dx=36\)
Đặt \(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=\frac{1}{2}x^2\end{matrix}\right.\)
\(\Rightarrow I=\frac{1}{2}x^2f\left(x\right)|^6_0-\frac{1}{2}\int\limits^6_0x^2.f'\left(x\right)dx\)
\(\Leftrightarrow36=18-\frac{1}{2}\int\limits^6_0x^2f'\left(x\right)dx\)
\(\Rightarrow\int\limits^6_0x^2f'\left(x\right)dx=-36\)
Lấy tích phân 2 vế giả thiết:
\(\int\limits^1_0\left(f'\left(x\right)\right)^2dx+4\int\limits^1_0f\left(x\right)dx=\int\limits^1_0\left(8x^2+4\right)dx=\frac{20}{3}\)
Xét \(I=\int\limits^1_0f\left(x\right)dx\)
Đặt \(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=x\end{matrix}\right.\)
\(\Rightarrow I=x.f\left(x\right)|^1_0-\int\limits^1_0x.f'\left(x\right)dx=2-\int\limits^1_0x.f'\left(x\right)dx\)
\(\Rightarrow\int\limits^1_0\left[f'\left(x\right)\right]^2dx+8-4\int\limits^1_0x.f'\left(x\right)dx=\frac{20}{3}\)
\(\Leftrightarrow\int\limits^1_0\left[f'\left(x\right)\right]^2dx-2\int\limits^1_02x.f'\left(x\right)dx+\int\limits^1_04x^2dx=\frac{20}{3}-8+\int\limits^1_04x^2dx=0\)
\(\Leftrightarrow\int\limits^1_0\left[\left[f'\left(x\right)\right]^2-2.2x.f'\left(x\right)+4x^2\right]dx=0\)
\(\Leftrightarrow\int\limits^1_0\left[f'\left(x\right)-2x\right]^2dx=0\Rightarrow f'\left(x\right)=2x\)
\(\Rightarrow f\left(x\right)=x^2+C\)
Do \(f\left(1\right)=2\Rightarrow2=1+C\Rightarrow C=1\)
\(\Rightarrow f\left(x\right)=x^2+1\Rightarrow\int\limits^1_0f\left(x\right)dx=\int\limits^1_0\left(x^2+1\right)dx=\frac{4}{3}\)
Đang học Lý mà thấy bài nguyên hàm hay hay nên nhảy vô luôn :b
\(I_1=\int\limits^1_0xf\left(x\right)dx\)
\(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=xdx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=\dfrac{1}{2}x^2\end{matrix}\right.\)
\(\Rightarrow\int xf\left(x\right)dx=\dfrac{1}{2}x^2f\left(x\right)-\dfrac{1}{2}\int x^2f'\left(x\right)dx\)
\(\Rightarrow\int\limits^1_0xf\left(x\right)dx=\dfrac{1}{2}x^2|^1_0-\dfrac{1}{2}\int\limits^1_0x^2f'\left(x\right)dx=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{1}{2}\int\limits^1_0\left[f'\left(x\right)\right]^2dx=\dfrac{3}{10}\Rightarrow\int\limits^1_0x^2f'\left(x\right)dx=\dfrac{3}{5}\)
Đoạn này hơi rối xíu, ông để ý kỹ nhé, nhận thấy ta có 2 dữ kiện đã biết, là: \(\int\limits^1_0\left[f'\left(x\right)\right]^2dx=\dfrac{9}{5}and\int\limits^1_0x^2f'\left(x\right)dx=\dfrac{3}{5}\) có gì đó liên quan đến hằng đẳng thức, nên ta sẽ sử dụng luôn
\(\int\limits^1_0\left[f'\left(x\right)+tx^2\right]^2dx=0\)
\(\Leftrightarrow\int\limits^1_0\left[f'\left(x\right)\right]^2dx+2t\int\limits^1_0x^2f'\left(x\right)dx+t^2\int\limits^1_0x^4dx=0\)
\(\Leftrightarrow\dfrac{9}{5}+\dfrac{6}{5}t+\dfrac{1}{5}t^2=0\) \(\left(\int\limits^1_0x^4dx=\dfrac{1}{5}x^5|^1_0=\dfrac{1}{5}\right)\)\(\)\(\Leftrightarrow t=-3\Rightarrow\int\limits^1_0\left[f'\left(x\right)-3x^2\right]^2dx=0\)
\(\Leftrightarrow f'\left(x\right)=3x^2\Leftrightarrow f\left(x\right)=x^3+C\)
\(\Rightarrow\int\limits^1_0f\left(x\right)dx=\int\limits^1_0x^3dx=\dfrac{1}{4}x^4|^1_0=\dfrac{1}{4}\)
P/s: Có gì ko hiểu hỏi mình nhé !
\(f\left(x\right)=\int sin^4xdx=\int\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2dx\)
\(=\frac{1}{4}\int\left(1-2cos2x+cos^22x\right)dx=\frac{1}{4}\int\left(\frac{3}{2}-2cos2x+\frac{1}{2}cos4x\right)dx\)
\(=\frac{1}{4}\left(\frac{3}{2}x-sin2x+\frac{1}{8}sin4x\right)+C\)
\(f\left(0\right)=0\Rightarrow\frac{1}{4}\left(0-0+0\right)+C=0\Rightarrow C=0\)
\(\Rightarrow\int\limits^{\frac{\pi}{2}}_0f\left(x\right)dx=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_0\left(\frac{3}{2}x-sin2x+\frac{1}{8}sin4x\right)dx\)
\(=\frac{1}{4}\left(\frac{3}{4}x^2+\frac{1}{2}cos2x-\frac{1}{32}cos4x\right)|^{\frac{\pi}{2}}_0\)
\(=\frac{3\pi^2-16}{64}\)
Bạn tham khảo:
Câu hỏi của T. Hữu Lộc - Toán lớp 12 | Học trực tuyến