CBSE Important Questions Class 12 Maths Chapter 8 – Application of Integrals

Class 12 Maths Chapter 8 Application of Integrals is part of the CBSE Class 12 Maths syllabus. It shows how definite integrals are used to find areas. Students learn to find the area under a simple curve, the area bounded by a curve and a line, and the area between two curves, using integration along either the x-axis or the y-axis.

These CBSE important questions for Class 12 Maths Chapter 8 cover the area under a parabola, the area of a circle and an ellipse using integration, the area between a curve and a line, the area of a region bounded by two curves, and the area of a triangle whose sides are given by straight lines. Every question has a clear step-by-step solution in simple English with figures, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Write the formula for the area under a curve. Find the area bounded by the curve y = x², the x-axis and the lines x = 1 and x = 3.

Answer: Formula: The area bounded by the curve y = f(x), the x-axis and the lines x = a and x = b is
Area = ∫ₐᵇ y dx = ∫ₐᵇ f(x) dx
Calculation:
Area = ∫₁³ x² dx
= [x³/3] from 1 to 3
= 27/3 − 1/3
= 26/3
= 26/3 square units

Q2. Find the area of the region bounded by the curve y² = 4x and the line x = 3.

Answer: The curve y² = 4x is a parabola opening to the right, and it is symmetric about the x-axis.
So we find the area above the x-axis and double it.
y = 2√x
Area = 2 ∫₀³ 2√x dx
= 4 ∫₀³ x^1/2 dx
= 4 [ (2/3) x^3/2 ] from 0 to 3
= (8/3) (3)^3/2
= (8/3)(3√3)
= 8√3 square units

Q3. Find the area of the circle x² + y² = a² using integration.

Answer: By symmetry, the circle has four equal quarters, so we find the area in the first quadrant and multiply by 4.
From the equation, y = √(a² − x²).
Area = 4 ∫₀ₐ √(a² − x²) dx
Using the standard result ∫√(a² − x²) dx = (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a):
= 4 [ (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) ] from 0 to a
= 4 [ 0 + (a²/2)(π/2) − 0 ]
= 4 × πa²/4
= πa² square units, which is the familiar formula for the area of a circle.

Q4. Find the area of the region bounded by the curve y = x² and the line y = 4.

Answer: First find where the curve meets the line:
x² = 4, so x = −2 and x = 2.
The region is symmetric about the y-axis, so we take twice the area from 0 to 2.
Area = 2 ∫₀² (4 − x²) dx
= 2 [ 4x − x³/3 ] from 0 to 2
= 2 [ 8 − 8/3 ]
= 2 × 16/3
= 32/3 square units

Q5. Find the area of the ellipse x²/a² + y²/b² = 1 using integration.

Answer: By symmetry we take the area in the first quadrant and multiply by 4.
From the equation, y = (b/a)√(a² − x²).
Area = 4 ∫₀ₐ (b/a)√(a² − x²) dx
= (4b/a) [ (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) ] from 0 to a
= (4b/a) [ (a²/2)(π/2) ]
= (4b/a)(πa²/4)
= πab square units
(Note that when a = b this becomes πa², the area of a circle, as expected.)

Q6. Find the area of the region bounded by the line y = 3x + 2, the x-axis and the lines x = −1 and x = 1.

Answer: The line crosses the x-axis where 3x + 2 = 0, that is x = −2/3. Below this point the line is below the x-axis, so we split the integral and take the modulus.
From −1 to −2/3 (area below the axis):
|∫ (3x + 2) dx| = |[3x²/2 + 2x]| = |(3/2)(4/9) − 4/3 − (3/2 − 2)| = 1/6
From −2/3 to 1 (area above the axis):
[3x²/2 + 2x] = (3/2 + 2) − (2/3 − 4/3) = 7/2 − (−2/3) = 25/6
Total area = 1/6 + 25/6 = 26/6 = 13/3 square units

Q7. Find the area between the two curves y = x² and y = x.

Answer: First find the points of intersection:
x² = x, so x(x − 1) = 0, giving x = 0 and x = 1.
Between 0 and 1, the line y = x lies above the parabola y = x².
Area = ∫₀¹ (upper − lower) dx
= ∫₀¹ (x − x²) dx
= [x²/2 − x³/3] from 0 to 1
= 1/2 − 1/3
= 1/6 square units

Q8. Find the area bounded by the curve y = sin x between x = 0 and x = 2π.

Answer: From 0 to π the curve is above the x-axis, and from π to 2π it is below. Area is always taken as positive, so we compute the two parts separately.
From 0 to π: ∫₀^π sin x dx = [−cos x] = −(−1) − (−1) = 2
From π to 2π: |∫^π^2π sin x dx| = |[−cos x]| = |−1 − 1| = 2
Total area = 2 + 2 = 4 square units
(If we had integrated straight through from 0 to 2π we would have got 0, because the two parts cancel.)

Q9. Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x² + y² = 32.

Answer: The line y = x meets the circle where 2x² = 32, so x = 4 (taking the positive value).
The region splits into two parts.
From x = 0 to 4, under the line:
∫₀⁴ x dx = [x²/2] = 8
From x = 4 to 4√2, under the circle:
∫₄^4√2 √(32 − x²) dx
Using the standard result, this evaluates to 4π − 8.
Total area = 8 + 4π − 8 = 4π square units

Q10. When do we integrate with respect to y instead of x?

Answer: We integrate with respect to y when the region is more naturally described in terms of horizontal strips, that is when the curve is given as x = f(y) or when the boundaries are horizontal lines.
Area = ∫ᶜᵈ x dy
For example, for the region bounded by x = y², the y-axis and the lines y = 1 and y = 3, it is much easier to write
Area = ∫₁³ y² dy = [y³/3] = 27/3 − 1/3 = 26/3 square units
than to try to split it into vertical strips.

Q11. Find the area of the triangle whose sides lie along the lines y = 2x + 1, y = 3x + 1 and x = 4.

Answer: First find the vertices by solving the lines in pairs:
y = 2x + 1 and y = 3x + 1 meet where 2x + 1 = 3x + 1, so x = 0 and y = 1, giving (0, 1).
y = 2x + 1 and x = 4 give (4, 9).
y = 3x + 1 and x = 4 give (4, 13).
Between x = 0 and 4, the line y = 3x + 1 is above y = 2x + 1.
Area = ∫₀⁴ [(3x + 1) − (2x + 1)] dx
= ∫₀⁴ x dx
= [x²/2] from 0 to 4
= 8 square units

Q12. Find the area bounded by the parabola y² = 4ax and its latus rectum.

Answer: For the parabola y² = 4ax, the latus rectum is the vertical line x = a.
The region is symmetric about the x-axis, so take twice the area above it.
y = 2√(ax)
Area = 2 ∫₀ₐ 2√(ax) dx
= 4√a ∫₀ₐ x^1/2 dx
= 4√a [ (2/3) x^3/2 ] from 0 to a
= (8√a/3) a^3/2
= 8a²/3 square units

Related Study Material on Extramarks

Q1.

Find the area of the region bounded by x² = y and y = |x|.

$\begin{array}{l} \end{array}$

Ans.

$\begin{array}{l} From y = |x| and x^{2} = y, we get \\ x = 1, -1 \\ The required area = 2 x [{ˆ«}_{0}^{1} y_{L} dx {ˆ«}_{0}^{1} y_{p} dx] \\ = 2 x [«_{0}^{1} xdx «_{0}^{1} x^{2} dx] \\ = 2 x \frac{x^{2}}{2} {\frac{x^{3}}{3}}_{0}^{1} \\ = 2 x [\frac{1}{2} \frac{1}{3}] \\ = 2 x [\frac{1}{6}] \\ = \frac{1}{3} sq . units \end{array}$

CBSE Important Questions Class 12 Maths Chapter 8 – Application of Integrals

Related Study Material on Extramarks

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