Ngân hàng bài tập
A
Cho hàm số $f(x)$ thỏa mãn $f(0)=4$ và $f'(x)=2\sin^2x+3$, $\forall x\in\mathbb{R}$. Khi đó $\displaystyle\displaystyle\int\limits_{0}^{\tfrac{\pi}{4}}f(x)\mathrm{d}x$ bằng
 | $\dfrac{\pi^2-2}{8}$ |
 | $\dfrac{\pi^2+8\pi-8}{8}$ |
 | $\dfrac{\pi^2+8\pi-2}{8}$ |
 | $\dfrac{3\pi^2+2\pi-3}{8}$ |
1 lời giải
Chọn phương án C.
$\begin{aligned}
f(x)&=\displaystyle\int f'(x)\mathrm{\,d}x=\displaystyle\int\big(2\sin^2x+3\big)\mathrm{\,d}x\\
&=\displaystyle\int\big((1-\cos2x)+3\big)\mathrm{\,d}x\\
&=\displaystyle\int\big(4-\cos2x\big)\mathrm{\,d}x\\
&=4x-\dfrac{1}{2}\sin2x+C.
\end{aligned}$
Vì $f(0)=4$ nên $4=4\cdot0-\dfrac{1}{2}\sin0+C\Leftrightarrow C=4$.
Vậy, $f(x)=4x-\dfrac{1}{2}\sin2x+4$. Do đó $$\displaystyle\int\limits_{0}^{\tfrac{\pi}{4}}f(x)\mathrm{d}x=\displaystyle\int\limits_{0}^{\tfrac{\pi}{4}}\left(4x-\dfrac{1}{2}\sin2x+4\right)\mathrm{d}x=\dfrac{\pi^2+8\pi-2}{8}.$$