Find the area bounded in the first quadrant between the ellipse x 2 16 + y 2 9 = 1 and the line 3 x + 4 y = 12

We have, x 2 16 + y 2 9 = 1 3 x + 4 y = 12 ⇒ x 4 + y 3 = 1 Required area = ∫ 0 4 3 4 16 - x 2 - 12 - 3 x 4 d x = 3 4 ∫ 0 4 16 - x 2 - 4 + x d x = 3 4 1 2 x 16 - x 2 + 16 2 sin - 1 x 4 - 4 x + x 2 2 0 4 = 3 4 0 + 8 sin - 1 4 4 - 4 × 4 + 16 2 - 0 = 3 4 8 × π 2 - 16 2 = 3 4 4 π - 8 = 3 π - 2 sq. units.

Find the area bounded in the first quadrant between the ellipse x 2 16 + y 2 9 = 1 and the line 3 x + 4 y = 12

We have, x 2 16 + y 2 9 = 1 3 x + 4 y = 12 ⇒ x 4 + y 3 = 1 Required area = ∫ 0 4 3 4 16 - x 2 - 12 - 3 x 4 d x = 3 4 ∫ 0 4 16 - x 2 - 4 + x d x = 3 4 1 2 x 16 - x 2 + 16 2 sin - 1 x 4 - 4 x + x 2 2 0 4 = 3 4 0 + 8 sin - 1 4 4 - 4 × 4 + 16 2 - 0 = 3 4 8 × π 2 - 16 2 = 3 4 4 π - 8 = 3 π - 2 sq. units.

Find the area of the region bounded by y = x and 2 x - 1 = 0 , y = 0 ( · stands for fractional part)

We have, y = x 2 x - 1 = 0 ⇒ x = 1 2 Equation of line O A is y = x . Now, required area = ∫ 0 1 2 x d x = x 2 2 0 1 2 = 1 8 sq . units

Let f x = tan x . Show that area bounded by y = f x , y = f c , x = 0 and x = a , 0 < c < a < π 2 is minimum when c = a 2 .

We have, y = tan x Required area A = ∫ 0 c f c - tan x d x + ∫ c a tan x - f c d x ⇒ A = ∫ 0 c f c d x - ∫ c a f c d x - ∫ 0 c tan x d x + ∫ c a tan x d x ⇒ A = f c c - 0 - f c a - c - ∫ 0 c tan x d x + ∫ c a tan x d x ⇒ A = 2 c tan c - a tan c - ∫ 0 c tan x d x + ∫ c a tan x d x Differentiate w.r.t. c , we get d A d c = 2 tan c + 2 c sec 2 c 2 tan c - a sec 2 c 2 tan c - tan c - tan c ⇒ d A d c = 2 c sec 2 c 2 tan c - a sec 2 c 2 tan c For critical points, d A d c = 0 ⇒ 2 c sec 2 c 2 tan c - a sec 2 c 2 tan c = 0 ⇒ 2 c - a = 0 ⇒ c = a 2 Hence, area if minimum when c = a 2 .

If A n is the area bounded by the curve y = tan x n and the lines x = 0 , y = 0 and x = π 4 . Then, prove that for n > 2 , A n + A n + 2 = 1 n + 1 and deduce 1 2 n + 2 < A n < 1 2 n - 2 .

We have, A n = ∫ 0 π 4 tan x n d x Since, 0 < tan x < 1 , when 0 < x < π 4 We have, 0 < tan x n + 1 < tan x n for each n ∈ N ⇒ ∫ 0 π 4 tan x n + 1 d x < ∫ 0 π 4 tan x n d x ⇒ A n + 1 < A n Now, for n > 2 A n + A n + 2 = ∫ 0 π 4 tan x n + tan x n + 2 d x ⇒ A n + A n + 2 = ∫ 0 π 4 tan x n 1 + tan 2 x d x ∵ ∫ f x n f ' x d x = f x n n + 1 ⇒ A n + A n + 2 = ∫ 0 π 4 tan x n sec 2 x d x ⇒ A n + A n + 2 = tan x n + 1 n + 1 0 π 4 ⇒ A n + A n + 2 = 1 n + 1 1 - 0 Since, A n + 2 < A n + 1 < A n ⇒ A n + A n + 2 < 2 A n ⇒ 1 n + 1 < 2 A n ⇒ 1 2 n + 2 < A n … i Also, for n > 2 , A n + A n < A n + A n - 2 ⇒ A n + A n < 1 n - 1 ⇒ 2 A n < 1 n - 1 ⇒ A n < 1 2 n - 2 … ii Combining i and ( ii ) , we get 1 2 n + 2 < A n < 1 2 n - 2