Integration is not just an algebraic operation — it has powerful geometric meaning. In this chapter, we use definite integrals to calculate areas of plane regions bounded by curves. This is one of the most practical applications of calculus in Class 12.
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Key Concepts
Area under a curve: The area bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b is:
Area = integral from a to b of f(x) dx
When f(x) is positive throughout [a, b], this gives a positive area. If f(x) is negative, the integral gives a negative value, so we take the absolute value.
Area between two curves: If f(x) ≥ g(x) on [a, b], the area between them is:
Area = integral from a to b of [f(x) - g(x)] dx
Area using horizontal strips (integrating along y): If x = g(y), the area between the curve and the y-axis from y = c to y = d is:
Area = integral from c to d of g(y) dy
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Key Formulas
Key formulas
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Worked Examples
Find the area bounded by y = x2, the x-axis, x = 0, and x = 3.
Area = integral from 0 to 3 of x2 dx = [x3/3] from 0 to 3 = 27/3 - 0 = 9 square units
Find the area bounded by y = sin x and the x-axis from x = 0 to x = pi.
Area = integral from 0 to pi of sin x dx = [-cos x] from 0 to pi = (-cos pi) - (-cos 0) = 1 + 1 = 2 square units
Find the area enclosed by the parabola y = x2 and the line y = x + 2.
First find intersections: x2 = x + 2 => x2 - x - 2 = 0 => (x-2)(x+1) = 0, so x = -1 and x = 2.
Area = integral from -1 to 2 of [(x + 2) - x2] dx = [x2/2 + 2x - x3/3] from -1 to 2
= (2 + 4 - 8/3) - (1/2 - 2 + 1/3) = 10/3 - (-7/6) = 20/6 + 7/6 = 27/6 = 9/2 square units
Find the area of the region bounded by the circle x2 + y2 = 4.
The circle has radius 2. Using the formula or integration:
Area in first quadrant = integral from 0 to 2 of √(4 - x2) dx = [x/2 √(4-x2) + 2 arcsin(x/2)] from 0 to 2 = pi
Total area = 4 pi = 4 pi square units (since r = 2, pi r2 = 4 pi)
Find the area bounded by y = |x + 1| and the x-axis from x = -3 to x = 1.
|x + 1| = -(x+1) for x < -1, and (x+1) for x ≥ -1.
Area = integral from -3 to -1 of [-(x+1)] dx + integral from -1 to 1 of (x+1) dx
= [-(x2/2 + x)] from -3 to -1 + [(x2/2 + x)] from -1 to 1
= (2 - 0) + (3/2 - (-1/2)) = 2 + 2 = 4 square units
Using integration, find the area of the smaller region bounded by the ellipse x2/9 + y2/4 = 1 and the line x/3 + y/2 = 1.
Intersections: at (3,0) and (0,2). Area = integral03 [2 √(1 - x2/9) - 2(1 - x/3)] dx
= (pi/2 × 3 × 2)/4 - 1/2 × 3 × 2/2 portion = (3 pi/2) area... by ellipse/triangle difference = (pi × 3 × 2)/4 - 3 = 3(pi - 2)/2 square units
Find the area between y = x3 and y = x from x = -1 to x = 1.
Note x3 ≤ x on [0,1] and x3 ≥ x on [-1,0].
By symmetry, Area = 2 × integral from 0 to 1 of (x - x3) dx = 2 × [x2/2 - x4/4] from 0 to 1 = 2 × (1/2 - 1/4) = 1/2 square unit
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Common mistakes
- Forgetting absolute value: When f(x) < 0 on part of [a, b], the integral can give a wrong (negative) area. Always split the interval at x-intercepts.
- Wrong subtraction order: Always subtract the lower curve from the upper curve: integral of [upper - lower], not [lower - upper].
- Not finding correct intersection points: Always solve for the limits of integration if not given explicitly.
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Summary
Area under a curve is computed using definite integrals. When a curve dips below the x-axis, take absolute values or split the integral. The area between two curves is the integral of their difference (upper minus lower). Standard shapes like circles and ellipses have formulas derivable by integration.