Answer to Question #110065 in Calculus for Nimra

Question

∫²∫¹sinx dx

Answer 1

$\smallint_0^2 \smallint_0^1 sin x dxdy = \smallint_0^2 [-cos x]^1_0 dy = \smallint_0^2(-cos 1+cos0)dy= \smallint_0^2(1-cos1)dy$

$= (1-cos1)\smallint_0^2dy = (1-cos1)[y]^2_0 = 2(1-cos1)=2(1-0.54)= 2(0.46)=0.92$