Cho \(f\left( x \right)\)có \(f\left( 0 \right) = 1\)và \(f\left( {\frac{\pi }{4}} \right) = \frac{\pi }{8}\) và \(f’\left( x \right) = \frac{{4m}}{\pi } + {\sin ^2}x\) (với \(m\) là tham số ). Tính \(\int\limits_0^\pi {f\left( x \right)} {\rm{dx}}\) ?
A. \( – \frac{\pi }{2} + \frac{{{\pi ^2}}}{8}\).
B. \( – 3 + \frac{\pi }{2}\).
C. \( – \frac{\pi }{2} + \frac{{{\pi ^2}}}{4}\).
D. \(1 – \frac{\pi }{2}\).
Lời giải
Chọn C
Ta có \(f\left( x \right) = \int {\left( {\frac{{4m}}{\pi } + {{\sin }^2}x} \right){\rm{dx}}} = \int {\left( {\frac{{4m}}{\pi } + \frac{{1 – \cos 2x}}{2}} \right)} {\rm{dx = }}\left( {\frac{{4m}}{\pi } + \frac{1}{2}} \right)x – \frac{1}{4}\sin 2x + C\)
Với \(\left\{ \begin{array}{l}f\left( 0 \right) = 1\\f\left( {\frac{\pi }{4}} \right) = \frac{\pi }{8}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}C = 1\\m = – \frac{3}{4}\end{array} \right. \Rightarrow f\left( x \right) = \left( {\frac{{ – 3}}{\pi } + \frac{1}{2}} \right)x – \frac{1}{4}\sin 2x + 1\)
Vậy \(\int\limits_0^\pi {f\left( x \right)} {\rm{dx = }}\int\limits_0^\pi {\left( {\left( {\frac{{ – 3}}{\pi } + \frac{1}{2}} \right)x – \frac{1}{4}\sin 2x + 1} \right)} {\rm{dx = }} – \frac{\pi }{2} + \frac{{{\pi ^2}}}{4}\).
Trả lời