Cho hàm số \(f\left( x \right)\). Biết \(f\left( 0 \right)=4\) và \({f}'\left( x \right)=2{{\sin }^{2}}x+1,\text{ }\forall x\in \mathbb{R}\), khi đó \(\int\limits_{0}^{\frac{\pi }{...

* Hướng dẫn giải

Ta có \(\int{{f}'\left( x \right)}\text{d}x=\int{\left( 2{{\sin }^{2}}x+1 \right)\text{d}x=\int{\left( 2-\cos 2x \right)}}\text{d}x=2x-\frac{1}{2}\sin 2x+C.\)

Suy ra \(f\left( x \right)=2x-\frac{1}{2}\sin 2x+C.\)

Vì \(f\left( 0 \right)=4\Rightarrow C=4\) hay \(f\left( x \right)=2x-\frac{1}{2}\sin 2x+4.\)

Khi đó: \(\int\limits_{0}^{\frac{\pi }{4}}{f\left( x \right)\text{d}x}=\int\limits_{0}^{\frac{\pi }{4}}{\left( 2x-\frac{1}{2}\sin 2x+4 \right)\text{d}x}\)

\(=\left( {{x}^{2}}+\frac{1}{4}\cos 2x+4x \right)\left| \begin{align} & \frac{\pi }{4} \\ & 0 \\ \end{align} \right.=\frac{{{\pi }^{2}}}{16}+\pi -\frac{1}{4}=\frac{{{\pi }^{2}}+16\pi -4}{16}.\)