Question

Given $f^{\prime }\left(x\right)=\frac{cosx}{x}$, $f\left(\frac{\pi }{2}\right)=a,f\left(\frac{3\pi }{2}\right)=b$. Find the value of the definite integral ${\int }_{\pi /2}^{3\pi /2}f\left(x\right)dx$.

Answer 1

$\frac{d}{dx}\left(xf\right(x\left)\right)=x{f}^{{}^{\prime }}\left(x\right)+f\left(x\right)$

$\Rightarrow f\left(x\right)=\frac{d}{dx}\left(xf\right(x\left)\right)-x{f}^{{}^{\prime }}\left(x\right)$

Integrating throughout,

$\int f\left(x\right)dx=xf\left(x\right)-\int x{f}^{{}^{\prime }}\left(x\right)dx$

ie, ${\int }_{\pi /2}^{3\pi /2}f\left(x\right)dx={xf\left(x\right)]}_{\pi /2}^{3\pi /2}-{\int }_{\pi /2}^{3\pi /2}\cos xdx$

$=\left\{\frac{3\pi }{2}f\right(\frac{3\pi }{2})-\frac{\pi }{2}f(\frac{\pi }{2}\left)\right\}-{sinx]}_{\pi /2}^{3\pi /2}=(\frac{3\pi }{2}b-\frac{\pi }{2}a)-\left(\right(-1)-1)=\frac{\pi }{2}(3b-a)+2$

$\therefore {\int }_{\pi /2}^{3\pi /2}f\left(x\right)dx=\frac{\pi }{2}(3b-a)+2$