Draw a rough sketch of the curve y = π 2 + 2 sin 2 x and find the area betweenx-axis, the curve and the ordinates x = 0 , x = π .

Given curves y = π 2 + 2 sin 2 x , x - axis, x = 0 and x = π x 0 π 6 π 4 π 3 π 2 2 π 3 3 π 4 5 π 6 π y 1 . 57 2 . 07 2 . 57 3 . 07 3 . 57 3 . 07 2 . 57 2 . 07 1 . 57 A rough sketch of the curves is drawn below: Required area = Area of the region A B C D O Area of the region enclosed by y = f ( x ) and x = a & x = b , b > a is given by ∫ a b f ( x ) d x ∴ Required area = ∫ 0 π y d x = ∫ 0 π π 2 + 2 sin 2 x d x = ∫ 0 π π 2 + 1 - cos 2 x d x = π 2 x + x - sin 2 x 2 0 π Since, if ∫ f ( x ) d x = F ( x ) + c , then ∫ f ( a x + b ) d x = 1 a F ( a x + b ) + C = π 2 2 + π - sin 2 x 2 - 0 = π 2 2 + π ∴ Required area = π 2 π + 2 sq . units .

Find the area bounded by the ellipse x 2 a 2 + y 2 b 2 = 1 and the ordinates x = a e and x = 0 , where b 2 = a 2 1 - e 2 and e < 1 .

The given ellipse can be drawn as : Note that the area of the region B O B ' R F S B is our required area. We know, area of the curve enclosed by y = f ( x ) and x = a & x = b , b > a is given by ∫ a b f ( x ) d x Area of the required region = 2 ∫ 0 a e y d x = 2 b a ∫ 0 a e a 2 - x 2 d x = 2 b a x 2 a 2 - x 2 + a 2 2 sin - 1 x a 0 a e = 2 b 2 a a e a 2 - a 2 e 2 + a 2 sin - 1 e = a b e 1 - e 2 + sin - 1 e