Application of Integrals: Complete Concepts and Practice

Big picture

The definite integral ∫ f(x) dx from a to b adds up the areas of very thin vertical strips under the curve y = f(x). That makes integration a tool for finding areas of curved regions, which ordinary geometry can't handle.

The board exam usually asks one long question here: sketch the region, decide the limits, set up the integral, and evaluate it. The curves are standard: straight lines, parabolas, circles and ellipses.

Learning goals

Find the area bounded by a curve, the x-axis and two vertical lines, or the y-axis and two horizontal lines.
Handle curves that lie below the x-axis.
Use symmetry to simplify area calculations.
Find the area enclosed by a circle, an ellipse, and a parabola with its latus rectum.
Find the area between two curves.

Core ideas

1. Area under a curve

Vertical strips. The area bounded by y = f(x) (with f(x) ≥ 0), the x-axis and the lines x = a and x = b is

A = ∫ y dx from a to b = ∫ f(x) dx from a to b

Horizontal strips. The area bounded by x = g(y), the y-axis and the lines y = c and y = d is

A = ∫ x dy from c to d

Choose whichever strip makes the integral simpler.

Example (line). Area bounded by y = 2x + 1, the x-axis, x = 1 and x = 3.

A = ∫ (2x + 1) dx from 1 to 3
  = [x² + x] from 1 to 3
  = (9 + 3) - (1 + 1) = 10 square units

(Check: the region is a trapezium with parallel sides 3 and 7 and width 2, area ½ × (3 + 7) × 2 = 10.)

Example (parabola). Area bounded by y = x², the x-axis, x = 1 and x = 3.

A = ∫ x² dx from 1 to 3 = [x³/3] from 1 to 3 = (27 - 1)/3 = 26/3 square units

2. Curves below the x-axis

If f(x) ≤ 0 on [a, b], the integral is negative, but area is always positive:

A = |∫ f(x) dx from a to b|

If the curve crosses the x-axis, split the interval at each crossing and add the absolute values.

Example. Area bounded by y = sin x and the x-axis from 0 to 2π. The curve is above the axis on [0, π] and below it on [π, 2π].

∫ sin x dx from 0 to π   = [-cos x] = 1 + 1 = 2
∫ sin x dx from π to 2π  = [-cos x] = -1 - 1 = -2
Area = 2 + |-2| = 4 square units

Integrating straight from 0 to 2π would wrongly give 0.

3. Using symmetry

Circles, ellipses and parabolas are symmetric about the axes. Find the area of one symmetric part and multiply.

Circle x² + y² = a². In the first quadrant, y = √(a² − x²). Using

∫ √(a² − x²) dx = (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + C,

Area of quarter = ∫ √(a² - x²) dx from 0 to a
                = [0 + (a²/2) sin⁻¹(1)] - [0]
                = (a²/2)(π/2) = πa²/4
Total area      = 4 × πa²/4 = πa²

For x² + y² = 16 (a = 4), the area is 16π, and the area in the first quadrant is 4π.

Ellipse x²/a² + y²/b² = 1. In the first quadrant, y = (b/a)√(a² − x²).

Area = 4 ∫ (b/a)√(a² - x²) dx from 0 to a
     = 4 (b/a)(πa²/4) = πab

Example. 9x² + 4y² = 36. Dividing by 36: x²/4 + y²/9 = 1, so a = 2, b = 3. Area = π × 2 × 3 = 6π square units.

Parabola y² = 4ax with its latus rectum x = a. The region is symmetric about the x-axis; the upper half is y = 2√(ax).

Area = 2 ∫ 2√a √x dx from 0 to a
     = 4√a × [(2/3) x^(3/2)] from 0 to a
     = 4√a × (2/3) a^(3/2) = 8a²/3

For y² = 4x (a = 1), the area cut off by the latus rectum x = 1 is 8/3 square units.

4. Area between two curves

If f(x) ≥ g(x) on [a, b], the area between the curves is

A = ∫ [f(x) − g(x)] dx from a to b (upper curve minus lower curve)

The limits are usually the x-coordinates of the points where the curves intersect.

Example. Area between y = x and y = x².

Intersection: x² = x gives x = 0 or x = 1. On [0, 1], x ≥ x² (for example, at x = ½: ½ > ¼).

A = ∫ (x - x²) dx from 0 to 1 = [x²/2 - x³/3] = 1/2 - 1/3 = 1/6 square units

Example. Area between y = x² and y = 2x. Intersection: x² = 2x gives x = 0 or 2.

A = ∫ (2x - x²) dx from 0 to 2 = [x² - x³/3] = 4 - 8/3 = 4/3 square units

5. Region bounded by a circle and a line (standard type)

Example. Area of the smaller region bounded by the circle x² + y² = 4 and the line x = 1.

The smaller region lies to the right of x = 1, between x = 1 and x = 2, symmetric about the x-axis.

A = 2 ∫ √(4 - x²) dx from 1 to 2
  = 2 [(x/2)√(4 - x²) + 2 sin⁻¹(x/2)] from 1 to 2
  = 2 [(0 + 2 × π/2) - (½ √3 + 2 × π/6)]
  = 2 [π - √3/2 - π/3]
  = 2 [2π/3 - √3/2]
  = 4π/3 - √3 square units

A reliable problem-solving routine

Sketch the curves and shade the required region; mark intersection points.
Decide on vertical or horizontal strips, whichever gives a simpler integrand.
Write the integrand as upper minus lower (or right minus left) and fix the limits.
Use symmetry where possible, and multiply at the end.
If part of the region is below the axis, take absolute values of separate integrals.
Evaluate carefully and write the answer with "square units".

Connections that matter

Integrals supplies the standard integrals, especially ∫ √(a² − x²) dx.
Conic sections (Class 11) gives the standard forms and shapes of the parabola, circle and ellipse.
Definite integrals as limit of a sum explains why the integral equals area.

Common mistakes

Getting a negative "area"; take the absolute value when the curve is below the axis.
Integrating across a crossing point, so positive and negative parts cancel.
Forgetting to multiply by the symmetry factor (2 or 4).
Subtracting the curves in the wrong order in area between curves.
Using wrong limits; find the intersection points first.

Revision checklist

I can find area under a curve with vertical or horizontal strips.
I can handle regions below the x-axis.
I can derive the areas of a circle (πa²), an ellipse (πab) and a parabola up to its latus rectum (8a²/3).
I can find the area between two curves.
I can sketch and set up any standard area problem.