cho f(x) có f(0)=0 và f'(x)=sinx^2.cos4x. Khi đó tích phân từ pi/2 đến 0 của f(x) bằng 10/09/2021 Bởi Clara cho f(x) có f(0)=0 và f'(x)=sinx^2.cos4x. Khi đó tích phân từ pi/2 đến 0 của f(x) bằng
Đáp án: $\dfrac{5}{36}$ Giải thích các bước giải: Ta có: $f'(x)=\sin^2x\cdot\cos4x$ $\to f(x)=\int f'(x)dx=\int \sin^2x\cdot\cos4xdx$ $\to f(x)=\int \dfrac12(1-\cos2x)\cdot\cos4xdx$ $\to f(x)=\dfrac12\int \cos4x-\cos2x\cdot\cos4xdx$ $\to f(x)=\dfrac12\int \cos4x-\dfrac12(\cos6x+\cos2x)dx$ $\to f(x)=\dfrac12\int \cos4x-\dfrac12\cos6x-\dfrac12\cos2xdx$ $\to f(x)=\dfrac{1}{2}\left(\dfrac{1}{4}\sin \left(4x\right)-\dfrac{1}{12}\sin \left(6x\right)-\dfrac{1}{4}\sin \left(2x\right)+C\right)$ Mà $f(0)=0$ $\to \dfrac{1}{2}\left(\dfrac{1}{4}\sin \left(4\cdot 0\right)-\dfrac{1}{12}\sin \left(6\cdot 0\right)-\dfrac{1}{4}\sin \left(2\cdot 0\right)+C\right)=0$ $\to C=0$ $\to f(x)=\dfrac{1}{2}\left(\dfrac{1}{4}\sin \left(4x\right)-\dfrac{1}{12}\sin \left(6x\right)-\dfrac{1}{4}\sin \left(2x\right)\right)$ $\to \int^0_{\dfrac{\pi}{2}}f(x)dx=\int^0_{\dfrac{\pi}{2}}\dfrac{1}{2}\left(\dfrac{1}{4}\sin \left(4x\right)-\dfrac{1}{12}\sin \left(6x\right)-\dfrac{1}{4}\sin \left(2x\right)\right)dx$ $\to \int^0_{\dfrac{\pi}{2}}f(x)dx=\dfrac{1}{2}\left(-\dfrac{1}{16}\cos \left(4x\right)+\dfrac{1}{72}\cos \left(6x\right)+\dfrac{1}{8}\cos \left(2x\right)\right)\Bigg|^0_{\dfrac{\pi}{2}}$ $\to \int^0_{\dfrac{\pi}{2}}f(x)dx=\dfrac{5}{36}$ Bình luận
Đáp án: $\dfrac{5}{36}$
Giải thích các bước giải:
Ta có:
$f'(x)=\sin^2x\cdot\cos4x$
$\to f(x)=\int f'(x)dx=\int \sin^2x\cdot\cos4xdx$
$\to f(x)=\int \dfrac12(1-\cos2x)\cdot\cos4xdx$
$\to f(x)=\dfrac12\int \cos4x-\cos2x\cdot\cos4xdx$
$\to f(x)=\dfrac12\int \cos4x-\dfrac12(\cos6x+\cos2x)dx$
$\to f(x)=\dfrac12\int \cos4x-\dfrac12\cos6x-\dfrac12\cos2xdx$
$\to f(x)=\dfrac{1}{2}\left(\dfrac{1}{4}\sin \left(4x\right)-\dfrac{1}{12}\sin \left(6x\right)-\dfrac{1}{4}\sin \left(2x\right)+C\right)$
Mà $f(0)=0$
$\to \dfrac{1}{2}\left(\dfrac{1}{4}\sin \left(4\cdot 0\right)-\dfrac{1}{12}\sin \left(6\cdot 0\right)-\dfrac{1}{4}\sin \left(2\cdot 0\right)+C\right)=0$
$\to C=0$
$\to f(x)=\dfrac{1}{2}\left(\dfrac{1}{4}\sin \left(4x\right)-\dfrac{1}{12}\sin \left(6x\right)-\dfrac{1}{4}\sin \left(2x\right)\right)$
$\to \int^0_{\dfrac{\pi}{2}}f(x)dx=\int^0_{\dfrac{\pi}{2}}\dfrac{1}{2}\left(\dfrac{1}{4}\sin \left(4x\right)-\dfrac{1}{12}\sin \left(6x\right)-\dfrac{1}{4}\sin \left(2x\right)\right)dx$
$\to \int^0_{\dfrac{\pi}{2}}f(x)dx=\dfrac{1}{2}\left(-\dfrac{1}{16}\cos \left(4x\right)+\dfrac{1}{72}\cos \left(6x\right)+\dfrac{1}{8}\cos \left(2x\right)\right)\Bigg|^0_{\dfrac{\pi}{2}}$
$\to \int^0_{\dfrac{\pi}{2}}f(x)dx=\dfrac{5}{36}$