Class 12 Maths Chapter 8: Application of Integrals MCQs with answers

Answer: (b) y dx

A vertical strip has height y = f(x) and width dx, so dA = y dx, and adding the strips from x = a to x = b gives A = ∫_a^b y dx. x dy is the area of a horizontal strip, used when the region is bounded by the y-axis.

Answer: (a) ∫₀² (y² + 1) dy

The region lies between the y-axis and the curve, so horizontal strips of length x = y² + 1 and width dy are added from y = 0 to y = 2: A = ∫₀² (y² + 1) dy. The limits 1 and 5 are the x-values of the end points, not the y-values; ∫₁⁵ √(x − 1) dx is the area between the curve and the x-axis, a different region.

Answer: (c) 2

A = ∫₀^{log 3} e^x dx = e^{log 3} − e⁰ = 3 − 1 = 2 square units. 3 forgets the lower limit; e³ − 1 takes 3 as the upper limit.

Answer: (d) 2

For 0 < x < 2 the line lies below the x-axis, so the integral is negative. Only its numerical value is the area: |−2| = 2 square units. An area is never negative.

Answer: (a) log 4

A = ∫₂⁸ dxx = log 8 − log 2 = log 4 square units. log 6 subtracts the limits inside the log; 38 comes from integrating as if the curve were 1x².

Answer: (c) 12

One arch of sin 2x runs from 0 to π2: ∫₀^π/2 sin 2x dx = [−12 cos 2x]₀^π/2 = 1. One arch of sin x has area ∫₀^π sin x dx = 2. The ratio is 12: the arch of sin 2x has the same height and half the width.

Answer: (d) 6

The area enclosed by x²a² + y²b² = 1 is πab. Here b = 2, so 2πa = 12π and a = 6. 3 takes b as 4; 12 drops b altogether; 36 is a².

Answer: (b) −∫_a^b f(x) dx

Below the x-axis every strip has y < 0, so ∫_a^b f(x) dx is negative and the area is its numerical value, |∫_a^b f(x) dx| = −∫_a^b f(x) dx. −∫_b^a f(x) dx equals ∫_a^b f(x) dx, which is negative.

Answer: (c) 812

x³ − 9x = x(x − 3)(x + 3) is positive on (−3, 0) and negative on (0, 3). ∫₀³ (x³ − 9x) dx = 814 − 812 = −814, and by symmetry the piece on (−3, 0) is +814. Area = 814 + 814 = 812. 0 is ∫₋₃³, where the two pieces cancel; 814 counts one piece only.

Answer: (a) 6

Split at x = 2. On [0, 2] the region is a triangle of area 2²2 = 2; on [2, k] it is a triangle of area (k − 2)²2. So 2 + (k − 2)²2 = 10, (k − 2)² = 16 and k = 6. 2 + 2√5 ignores the triangle on [0, 2].