Chapter 7 Application of Definite Integration Miscellaneous Exercise 7 Solutions

(a) sq units

(c) sq units

(c) sq units

(b) sq units

(c) 25π sq units

sq units

49π sq units

sq units

234 sq units

sq units

True

False

True

False

True

= c2 log() = c2. log 2 sq units.

For finding the points of intersection of the two parabolas, we equate the values of y2 from their equations. From the equation x2 = 7y, y2 = ∴ = 7x ∴ x4 = 343x ∴ x4 – 343x = 0 ∴ x(x3 – 343) = 0 ∴ x = 0 or x3 = 343, i.e. x = 7 When x = 0, y = 0 When x = 7, 7y = 49 ∴ y = 7 ∴ the points of intersection are O(0, 0) and A(7, 7) Required area = area of the region OBACO = (area of the region ODACO) – (area of the region ODABO) Now, area of the region ODACO = area under the parabola y2 = 7x i.e. y = √7 √x Area of the region ODABO = Area under the parabola x2 = 7y i.e. y = ∴ required area = sq units.

By the symmetry of the parabola, the required area is twice the area of the region OABCO Now, the area of the region OABCO

By the symmetry of the ellipse, the required area of the ellipse is 4 times the area of the region OPQO. For the region OPQO, the limits of integration are x = 0 and x = 4.

Required area = , where y = x2 = = = 21 sq units.

Given the equation of the parabola is y2 = 25x ∴ y = 5√x …… [∵ IIn first quadrant, y > 0] Required area = area of the region OQRPO = 2(area of the region ORPO)