Question

please anybody solve question in image ​

Answer 1

Answer:

Option (iv) 2 is correct

Step-by-step explanation:

Given :

A function of x,y : Sin(x+y)

To find :

Integration of following:

$\int_{0}^{\pi/2}{ \right.\int_{0}^{\pi/2}sin(x+y)dxdy$

Solution:

$\int_{0}^{\pi/2}sin(x+y)dx = -cos(\frac{\pi}{2}+y)-[ -cos(0+y) ]\\=-(-siny) )+cosy= siny + cosy$

Thus

$\int_{0}^{\pi/2}{ \right.\int_{0}^{\pi/2}sin(x+y)dxdy$

$= \int_{0}^{\pi/2}(siny + cosy)dy\\\\=( -cos \pi/2 -(- cos0)} + (sin \pi/2-sin 0)\\\\=-0+1+1-0\\=2$

Answer:

$\int_{0}^{\pi/2}{ \right.\int_{0}^{\pi/2}sin(x+y)dxdy =2$

Formulas used:

$(1) \int cosx dx =sinx\\ and \, \,(2) \, \int sinx dx=-cos x$