Question

$\int\limits^b_a cos x dx ,Obtain the given integrals as the limit of a sum.$

Answer 1

HELLO DEAR,


$\sf{\int\limits^b_a cosx\,dx}$

we know the integral of cos.dx = sinx


so, $\sf{\int\limits^b_acosx.dx = [sinx]^b_a}$

$\sf{= [sinb - sina]}$


I HOPE ITS YOU DEAR,
THANKS

Answer 2

we have to get the value of $\int\limits^b_a{cosx}\,dx$

we know, $\int{cosx}\,dx=\left[sinx\right]+C$

here,
$\int\limits^b_a{cosx}\,dx\\\\\\=\left[sinx\right]^b_a\\\\\\=\left[sinb-sina\right]\\\\\\=2cos\frac{(b+a)}{2}.sin\frac{(b-a)}{2}$