Cho hàm số \(f\left( x \right)\) có đạo hàm liên tục trên đoạn \(\left[ 0;\frac{\pi }{4} \right]\) thỏa mãn \(f\left( 0 \right)=0, \int\limits_{\text{0}}^{\frac{\pi }{\text{4}}}{{{...

* Hướng dẫn giải

Đặt \(u=f\left( x \right)\Rightarrow \text{d}u={f}'\left( x \right)\text{d}x,\text{d}v=\sin 2x\text{d}x\Rightarrow v=-\frac{1}{2}\cos 2x.\)

Do đó: \(\frac{1}{2}=\int\limits_{\text{0}}^{\frac{\pi }{\text{4}}}{\sin 2x.f\left( x \right)\text{d}x}=\left. \left( -\frac{f\left( x \right)}{2}\cos 2x \right) \right|_{0}^{\frac{\pi }{4}}+\int\limits_{\text{0}}^{\frac{\pi }{\text{4}}}{\frac{1}{2}\cos 2x.{f}'\left( x \right)\text{d}x}\)

\(\Rightarrow \int\limits_{\text{0}}^{\frac{\pi }{\text{4}}}{\left( \cos 2x \right).{f}'\left( x \right)\text{d}x}=1\Rightarrow \int\limits_{\text{0}}^{\frac{\pi }{\text{4}}}{\left( 2\cos 2x \right).{f}'\left( x \right)\text{d}x}=2\Rightarrow {f}'\left( x \right)=2\cos 2x.\)

\(\Rightarrow f\left( x \right)=\sin 2x+C.\) Mà \(f\left( 0 \right)=0\) nên \(C=0\Rightarrow f\left( x \right)=\sin 2x.\)

\(\int\limits_{\text{0}}^{\frac{\pi }{\text{4}}}{f\left( x \right)\text{d}x}=\int\limits_{\text{0}}^{\frac{\pi }{\text{4}}}{\sin 2x\text{d}x}=\left. \left( -\frac{1}{2}\cos 2x \right) \right|_{0}^{\frac{\pi }{4}}=\frac{1}{2}.\)