Sketch the region bounded by the curve y = 1 2 2 - 3 x - 2 x 2 below the line y = x + 1 and above the x - a x i s , also find its area.

Given curves are y = 1 2 2 - 3 x - 2 x 2 = - 1 2 2 x - 1 x + 2 & y = x + 1 Sketch ∴ Required Area = Area of △ O A B + ∫ 0 1 2 1 2 2 - 3 x - 2 x 2 d x = 1 2 × 1 × 1 + 1 2 2 x - 3 x 2 2 - 2 x 3 3 0 1 2 = 1 2 + 1 2 1 - 3 8 - 1 12 = 37 48 s q . u n i t

Find the area bounded by the curve y = 4 - x 2 , x 2 + y 2 - 4 x = 0 and x - a x i s .

Given curves are y = 4 - x 2 which is semicircle lies above the x - a x i s whose center is 0 , 0 & radius is 2 and x 2 + y 2 - 4 x = 0 is equation of circle whose Centre is 2 , 0 & radius is 2 . Combine Sketch Point of intersection of both the circles, 4 - x 2 = 4 x - x 2 ⇒ x = 1 ∴ R e q u i r e d A r e a = ∫ 0 1 4 x - x 2 d x + ∫ 1 2 4 - x 2 d x = ∫ 0 1 4 - x - 2 2 d x + ∫ 1 2 4 - x 2 d x = x - 2 2 4 - x - 2 2 + 4 2 sin - 1 x - 2 2 0 1 + x 2 4 - x 2 + 4 2 sin - 1 x 2 1 2 = - 1 2 3 + 2 - π 6 - 0 + 2 - π 2 + 2 . π 2 - 1 2 3 + 2 π 6 = - 3 + 4 π 3 = 4 π 3 - 3 s q . u n i t

Prove that the ratio in which the area of the semi-circle, y = 5 - x 2 is divided by the normal to it at - 1 , 2 is π + 2 sin - 1 1 5 : π - 2 sin - 1 1 5 .

Given curve y = 5 - x 2 ⇒ d y d x = - 2 x 2 5 - x 2 ⇒ d y d x - 1 , 2 = 1 2 ∴ slope of Normal at - 1 , 2 is - 2 ⇒ Equation of Normal at - 1 , 2 is y - 2 = - 2 x + 1 ⇒ y = - 2 x Sketch Area of portion O A B = ∫ - 5 - 1 5 - x 2 d x + ∫ - 1 0 - 2 x d x = x 2 5 - x 2 + 5 2 sin - 1 x 5 - 5 - 1 - x 2 - 1 0 = - 1 2 4 + 5 2 sin - 1 - 1 5 - 5 2 × - π 2 - 0 - 1 = - 5 2 sin - 1 1 5 + 5 4 π sq . units Area of semi-circle = 5 π 2 sq.units Required ratio = 5 π 2 - 5 π 4 - 5 2 sin - 1 1 5 5 π 4 - 5 2 sin - 1 1 5 = 5 π 4 + 5 2 sin - 1 1 5 5 4 π - 2 sin - 1 1 5 = π + 2 sin - 1 1 5 π - 2 sin - 1 1 5 Hence, proved.