Class 12 Maths Chapter 8: Application of Integrals worksheet with answer key

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: (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) 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: (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.

4x − x² = x(4 − x) is zero at x = 0 and x = 4 and positive between. A = ∫₀⁴ (4x − x²) dx = [2x² − x³3]₀⁴ = 32 − 643 = 323 square units.

• finds the limits x = 0 and x = 4 and sets up ∫₀⁴ (4x − x²) dx (1)
• area 323 square units (1)

The area enclosed by the ellipse is πab. With b = 6 − a, the area is A = πa(6 − a) = π[9 − (a − 3)²]. This is greatest when a = 3 (equivalently dA/da = π(6 − 2a) = 0, with d²A/da² = −2π < 0, Chapter 6), so a = b = 3: the ellipse is the circle x² + y² = 9, and the greatest area is 9π square units.

• area πa(6 − a), from πab with b = 6 − a (1)
• greatest at a = b = 3, the circle x² + y² = 9, area 9π square units (1)

x² − 2x = x(x − 2) is zero at x = 0 and x = 2 and negative between, so the region lies below the x-axis. ∫₀² (x² − 2x) dx = 83 − 4 = −43; the area is its numerical value, 43 square units.

• limits 0 and 2 and the integral −43 (1)
• takes the numerical value: area 43 square units (1)

sin x + cos x = 0 where tan x = −1, that is at x = 3π4; the curve is above the x-axis on (0, 3π4) and below on (3π4, π). An anti derivative is sin x − cos x. ∫₀^3π/4 = (1√2 + 1√2) − (0 − 1) = √2 + 1. ∫_3π/4^π = (0 + 1) − √2 = 1 − √2, numerical value √2 − 1. Area = (√2 + 1) + (√2 − 1) = 2√2 square units.

• zero at x = 3π4 and the split (1)
• the two integrals √2 + 1 and 1 − √2 (1)
• area 2√2 square units (1)

y = x³ − 3x² + 2x is positive on (0, 1) and negative on (1, 2). An anti derivative is F(x) = x⁴4 − x³ + x². ∫₀¹ y dx = F(1) − F(0) = 14; ∫₁² y dx = F(2) − F(1) = 0 − 14 = −14. Area = 14 + 14 = 12 square unit.

• finds the sign change at x = 1 and splits the integral (1)
• evaluates the pieces 14 and −14 (1)
• area 12 square unit (1)

The curve meets the y-axis at (0, 1) and the line y = e at (1, e). (a) On the curve x = log y, so A = ∫₁^e log y dy = [y log y − y]₁^e = (e − e) − (0 − 1) = 1 square unit. (b) The rectangle with corners (0, 0) and (1, e) has area e; the area under y = e^x from x = 0 to x = 1 is e − 1; the difference is e − (e − 1) = 1 square unit.

• limits y = 1 to y = e with x = log y (1)
• integrates log y by parts to 1 (1)
• rectangle e less e − 1 gives 1 (1)