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Câu hỏi của T. Hữu Lộc - Toán lớp 12 | Học trực tuyến
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 \(f\left(1\right)=2\Rightarrow2=1+C\Rightarrow C=1\)
\(\Rightarrow f\left(x\right)=x^2+1\Rightarrow\int\limits^1_0f\left(x\right)dx=\int\limits^1_0\left(x^2+1\right)dx=\frac{4}{3}\)
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 \(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=\frac{1}{2}x^2\end{matrix}\right.\)
\(\Rightarrow I=\frac{1}{2}x^2f\left(x\right)|^6_0-\frac{1}{2}\int\limits^6_0x^2.f'\left(x\right)dx\)
\(\Leftrightarrow36=18-\frac{1}{2}\int\limits^6_0x^2f'\left(x\right)dx\)
\(\Rightarrow\int\limits^6_0x^2f'\left(x\right)dx=-36\)
Chọn \(f\left(x\right)=x^3+ax^2+bx+c\)
\(2f\left(x^2\right)+f'\left(x\right)=2x^6+7x^2+2\)
\(\Leftrightarrow2x^6+2ax^4+2bx^2+c+3x^2+2ax+b=2x^6+7x^2+2\)
\(\Leftrightarrow2ax^4+\left(2b+3\right)x^2+2ax+b+c=7x^2+2\)
Đồng nhất 2 vế ta được: \(\left\{{}\begin{matrix}a=0\\2b+3=7\\b+c=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=c=0\\b=2\end{matrix}\right.\)
\(\Rightarrow f\left(x\right)=x^3+2x\Rightarrow f\left(1\right)=3\)