Application of Integrals

In Class 12 Calculus, one of the most useful geometric applications of the definite integral is calculating the area of a plane region bounded by a curve. If y = f(x) is a continuous curve defined on the interval [a, b] and f(x) ≥ 0 for every x in [a, b], then the region bounded by the curve, the x-axis, and the ordinates x = a and x = b has area equal to the definite integral of f(x) from a to b.

This works because a definite integral is itself defined as the limit of a sum of areas of thin rectangular strips of width Δx and height f(x), taken across the interval — exactly the elementary area of the region under the curve.

Worked Example: Find the area of the region bounded by the curve y = x², the x-axis, and the lines x = 0 and x = 2.

Since y = x² ≥ 0 on [0, 2], the required area is:

A = ∫₀² x² dx = [x³/3]₀² = 8/3 − 0 = 8/3 square units.

If the curve lies below the x-axis on [a, b] (i.e., f(x) ≤ 0), the definite integral ∫f(x)dx comes out negative, but area is always a non-negative quantity — so we take the absolute value (modulus) of the integral: Area = |∫ₐᵇ f(x) dx|.

Worked Example: Find the area bounded by y = sin x, the x-axis, and the lines x = π and x = 2π.

Here sin x ≤ 0 throughout [π, 2π]. ∫_π^2π sin x dx = [−cos x]_π^2π = (−cos 2π) − (−cos π) = −1 − 1 = −2. Since area cannot be negative, the required area is |−2| = 2 square units.

Sometimes a curve is more conveniently expressed as x = g(y) rather than y = f(x) — for example, a sideways-opening parabola like x = y². In such cases it is easier to slice the region into thin horizontal strips of width Δy rather than vertical strips.

If the curve x = g(y) is continuous and g(y) ≥ 0 for y between c and d, the area of the region bounded by the curve, the y-axis, and the horizontal lines y = c and y = d is given by the definite integral of g(y) with respect to y.

Worked Example: Find the area of the region bounded by the curve x = y², the y-axis, and the lines y = 1 and y = 4.

A = ∫₁⁴ x dy = ∫₁⁴ y² dy = [y³/3]₁⁴ = 64/3 − 1/3 = 63/3 = 21 square units.

As with the x-axis case, if g(y) is negative over part of [c, d] (the curve lies to the left of the y-axis), the corresponding part of the integral is taken in absolute value before adding to the total area.

Two standard curves that appear repeatedly in this chapter are the circle x² + y² = a² and the ellipse x²/a² + y²/b² = 1. Both are symmetric about both coordinate axes, so it is efficient to compute the area of one quadrant using integration and then multiply by 4.

Circle: For x² + y² = a², solving for y in the first quadrant gives y = √(a² − x²). The area of the full circle is four times the first-quadrant area:

A = 4∫₀ᵃ √(a² − x²) dx = 4·[ (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) ]₀ᵃ = 4·(a²π/4) = πa².

Worked Example: Find the area enclosed by the circle x² + y² = 4.

Here a = 2, so the area = πa² = π(2)² = 4π square units.

Ellipse: For x²/a² + y²/b² = 1, solving for y in the first quadrant gives y = (b/a)√(a² − x²). By the same symmetry argument:

A = 4∫₀ᵃ (b/a)√(a² − x²) dx = (4b/a)·(a²π/4) = πab.

Worked Example: Find the area of the ellipse x²/9 + y²/4 = 1.

Here a = 3, b = 2, so the area = πab = π(3)(2) = 6π square units.

A very common problem type is finding the area of the region enclosed between a curve (often a parabola, circle, or ellipse) and a straight line. The general method has three steps: (1) solve the two equations simultaneously to find the points of intersection, (2) reason out which curve lies above (or to the right of) the other over the interval, and (3) integrate the difference between the two functions over that interval.

Worked Example: Find the area of the region bounded by the parabola y² = x and the line y = x − 2.

Since both curves are naturally expressed in terms of y (x = y² and x = y + 2), it is convenient to integrate with respect to y. Substituting x = y² into x = y + 2 gives y² = y + 2, i.e., y² − y − 2 = 0, which factors as (y − 2)(y + 1) = 0. So the curves intersect at y = −1 and y = 2.

For y between −1 and 2, the line x = y + 2 lies to the right of the parabola x = y². So the area is:

A = ∫₋₁² [(y + 2) − y²] dy = [y²/2 + 2y − y³/3]₋₁²

At y = 2: 4/2 + 4 − 8/3 = 2 + 4 − 8/3 = 6 − 8/3 = 10/3.
At y = −1: 1/2 − 2 − (−1/3) = 1/2 − 2 + 1/3 = −7/6.

A = 10/3 − (−7/6) = 10/3 + 7/6 = 20/6 + 7/6 = 27/6 = 9/2 square units.

The idea of the previous section extends directly to two general curves y = f(x) and y = g(x). If f(x) ≥ g(x) throughout the interval [a, b], and the curves meet at x = a and x = b, then the enclosed area is the integral of their difference.

Worked Example: Find the area of the region enclosed between the parabola y = x² and the line y = x.

Points of intersection: x² = x ⟹ x² − x = 0 ⟹ x(x − 1) = 0 ⟹ x = 0, 1. On (0, 1), the line y = x lies above the parabola y = x² (check at x = 0.5: line gives 0.5, parabola gives 0.25).

A = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = (1/2 − 1/3) − 0 = 1/6 square unit.

When two curves intersect at more than two points, or when one curve is above the other for part of the interval and below it for another part, the interval must be split at each intersection point, and the absolute value of the (outer − inner) integral is added up over each sub-interval — otherwise areas above and below would wrongly cancel.

Two techniques make area problems far quicker to solve correctly: careful handling of sign changes, and exploiting symmetry.

Sign convention: A definite integral ∫f(x)dx gives a signed value — positive when the curve is above the x-axis, negative when it is below. Since geometric area can never be negative, whenever a curve crosses the x-axis within the interval of integration, the interval must be split at each crossing point, and the absolute value of each piece added together.

Worked Example: Find the area bounded by the curve y = x³ − x and the x-axis between x = −1 and x = 1.

y = x³ − x = x(x − 1)(x + 1), which is zero at x = −1, 0, 1. Testing x = −0.5 gives y = 0.375 > 0, and testing x = 0.5 gives y = −0.375 < 0. So the curve is above the axis on [−1, 0] and below it on [0, 1].

∫(x³ − x)dx = x⁴/4 − x²/2. On [−1, 0]: value at 0 is 0; value at −1 is 1/4 − 1/2 = −1/4. So ∫₋₁⁰ = 0 − (−1/4) = 1/4.
On [0, 1]: value at 1 is 1/4 − 1/2 = −1/4; value at 0 is 0. So ∫₀¹ = −1/4, whose absolute value is 1/4.

Total area = 1/4 + 1/4 = 1/2 square unit. (Notice the two pieces came out equal — this is because y = x³ − x is an odd function, symmetric about the origin.)

Symmetry: If a curve is symmetric about the y-axis (an even function) or about the origin (an odd function), the area of one symmetric part can be computed and simply doubled — this was already used for the circle and ellipse, and again here for y = x³ − x.