Discussion Forum

Match the following

Answer:

Explanation:

$I. \int_{-1}^{1} x|x|dx=0$    $[\because x|x|$ is an odd function]

II.Let I= $\int_{0}^{\pi/2} \left[1+\left( \log \frac{4+3 \sin x}{4+3 \cos x}\right)\right]dx$

  $I= \int_{0}^{\pi/2}\left(1+ \left[ \log \frac{4+3 \sin \left(\frac{\pi}{2}-x\right)}{4+3 \cos \left( \frac{\pi}{2}-x\right)})\right]\right)dx$

$I= \int_{0}^{\pi/2}\left(1+ \left[ \log \frac{4+3 \cos x}{4+3 \sin x}\right]\right)dx$

 $\Rightarrow$     $2 I= \int_{0}^{\pi/2}\left(2+  \log \frac{4+3 \sin x}{4+3 \cos x}+\log \frac{4+3 \cos x}{4+3 \sin x}\right)dx$

    $\Rightarrow$     $2 I= \int_{0}^{\pi/2}\left(2+  \log \frac{(4+3 \sin x)(4+3 \cos x)}{(4+3 \cos x)(4+3 \sin x)}\right)$

 $\Rightarrow$    $2 I= \int_{0}^{\pi/2}(2+ \log 1)dx=\int_{0}^{\pi/2} 2dx= \pi$

 $\Rightarrow$    $I= \frac{\pi}{2}$

 III.    $\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx$

IV.   $\int_{-a}^{a} f(x) dx= \int_{0}^{a} f(x) dx+\int_{0}^{a} f(-x) dx$

   =  $\int_{0}^{a} [f(x) +f(-x)]dx$