Evaluate the integral?

intRint siny/y dA, R;the region bounded by y=x and y=sqrtx

1 Answer
Jun 27, 2018

= 1 - sin 1

Explanation:

intint_R siny/y dA

= int_0^1 dx int_x^sqrtx dy qquad siny/y

= int_0^1 dy int_(y^2)^y dx qquad siny/y

The second one looks more promising:

= int_0^1 dy qquad [ xsiny/y]_(x=y^2)^y

= int_0^1 dy qquad siny -y siny

= [(y - 1) cosy - siny]_0^1 = 1 - sin 1

graph{(y - sqrtx)(y-x)=0 [-1.502, 1.5, -0.25, 1.517]}