Application of Integrals

In this chapter: area under a curve using vertical strips, area between a curve and the y-axis using horizontal strips, regions below the x-axis, areas bounded by circles, ellipses and parabolas, and area between two curves.

Area under a curve

Divide the region between the curve y = f(x), the x-axis and the lines x = a, x = b into thin vertical strips. A strip at x has height y and width dx, so its area is y dx, the elementary area. Adding the strips from a to b gives

If it is more convenient to express x in terms of y, use horizontal strips of length x and width dy. The area bounded by x = g(y), the y-axis and the lines y = c, y = d is

Area is always positive. If the curve lies below the x-axis between a and b, the integral comes out negative, and the area is its absolute value. If the curve crosses the axis inside [a, b], split the interval at the crossing points and add the absolute values of the pieces.

Worked example: Find the area bounded by y = sin x and the x-axis between x = 0 and x = 2π.
Solution: sin x ≥ 0 on [0, π] and ≤ 0 on [π, 2π]. ∫₀^π sin x dx = [−cos x]₀^π = 2 and ∫_π^2π sin x dx = −2. Area = 2 + |−2| = 4 square units. The single integral from 0 to 2π gives 0, which is the signed value, not the area.

Standard regions

Circle and ellipse

Use symmetry. For the circle x² + y² = a², the first-quadrant part is y = √(a² − x²), so the area is 4∫₀^a √(a² − x²) dx = 4 · πa²/4 = πa², using the standard integral for √(a² − x²).

Worked example: Find the area enclosed by the ellipse x²/9 + y²/4 = 1.
Solution: In the first quadrant, y = (2/3)√(9 − x²). Area = 4 ∫₀³ (2/3)√(9 − x²) dx = (8/3) [ (x/2)√(9 − x²) + (9/2) sin⁻¹(x/3) ]₀³ = (8/3)(9/2)(π/2) = 6π square units. This agrees with the general result πab = π(3)(2).

Parabola

The parabola y² = 4ax is symmetric about the x-axis, so work with the upper half y = 2√(ax) and double it.

Worked example: Find the area of the region bounded by y² = 4x and the line x = 1.
Solution: Area = 2 ∫₀¹ 2√x dx = 4 · [ (2/3) x^3/2 ]₀¹ = 8/3 square units. In general, the area bounded by y² = 4ax and its latus rectum x = a is 8a²/3.

Use this table to check answers, not to replace working. In a board answer the integral must be shown.

Area between two curves

If y = f(x) lies above y = g(x) on [a, b], a vertical strip has height f(x) − g(x), so

Area between two curves was removed from the rationalised NCERT book, but it is still in the JEE and MHT‑CET syllabus and is asked often. The limits a and b are usually the x-coordinates of the points where the curves meet, found by solving the two equations together. If the curves cross inside the interval, the upper and lower curves swap, and the interval must be split there.

Worked example: Find the area enclosed between y = x and y = x².
Solution: The curves meet where x² = x, that is at x = 0 and x = 1. For 0 < x < 1, x > x² (for example, 1/2 > 1/4), so the line is the upper curve. Area = ∫₀¹ (x − x²) dx = 1/2 − 1/3 = 1/6 square unit.

Sometimes horizontal strips are simpler. For the region between the parabola x = y² and the line x = 4, the right boundary is x = 4 and the left boundary is x = y² for −2 ≤ y ≤ 2, giving A = ∫₋₂² (4 − y²) dy = 16 − 16/3 = 32/3.

How to approach any area question

• Sketch the curves and shade the required region. Even a rough sketch prevents most errors.
• Find the points of intersection to get the limits.
• Decide between vertical strips (integrate in x) and horizontal strips (integrate in y); choose whichever gives a single integral.
• Use symmetry to reduce the work, then multiply.

Common mistakes: (1) Integrating straight across a point where the curve crosses the x-axis, so positive and negative parts cancel. (2) Writing lower curve minus upper curve and reporting a negative area. (3) Using x-limits with a dy integral, or the other way round. (4) Forgetting to multiply by 2 or 4 after using symmetry. (5) Finding intersection points wrongly because only one of the equations was used.

JEE and MHT‑CET focus

• Areas bounded by a parabola and a line, two parabolas, or a circle and a line.
• Area of an ellipse or a part of it cut off by a chord.
• Regions involving |x| and other piecewise curves, found by splitting the interval.
• Choosing horizontal strips when the curve is given as x = g(y).
• Sketching the region quickly from the equations; almost every question needs it.

Practice questions