Applications of Integrals

Area Under Simple Curves

Understand

If \(f(x)\ge0\) on \([a,b]\), the area of the region bounded by \(y=f(x)\), the \(x\)-axis and the lines \(x=a,\ x=b\) is

\[A=\int_a^bf(x)\,dx=\int_a^by\,dx.\]

24xy
Area under \(y=x^2\) from \(x=0\) to \(x=2\) is \(\int_0^2x^2dx=\frac83\).

Think of the region as made of thin vertical strips of height \(y\) and width \(dx\); integration adds them up. Similarly, with horizontal strips, the area between \(x=g(y)\), the \(y\)-axis and \(y=c,\ y=d\) is \(\int_c^dx\,dy\).

Regions below the axis

If \(f(x)\le0\), \(\int_a^bf\) is negative; the area is \(\left|\int_a^bf\right|\). If \(f\) changes sign, split the interval at its zeros and add the absolute values.

Standard curves

2−22xy
Circle \(x^2+y^2=4\): area \(=4\int_0^2\sqrt{4-x^2}\,dx=4\pi\).
  • Circle \(x^2+y^2=a^2\): \(A=4\int_0^a\sqrt{a^2-x^2}\,dx=\pi a^2\).
  • Ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\): \(A=4\cdot\frac ba\int_0^a\sqrt{a^2-x^2}\,dx=\pi ab\).
  • Parabola \(y^2=4ax\) and the line \(x=a\) (latus rectum): \(A=2\int_0^a2\sqrt{ax}\,dx=\frac{8a^2}{3}\).

Method

  1. Sketch the curve and shade the region.
  2. Decide vertical (dx) or horizontal (dy) strips.
  3. Write the limits from the sketch; use symmetry.
  4. Integrate and state the area in square units.

Key Concepts

  • Area = ∫ y dx (or ∫ x dy) with the correct limits.
  • Area is always positive: take absolute values below the axis.
  • Use symmetry for circles, ellipses and parabolas.

Formula Bank

Area under a curve

\[A=\int_a^b y\,dx\quad\text{or}\quad A=\int_c^d x\,dy\]

Circle and ellipse

\[\int_0^a\sqrt{a^2-x^2}\,dx=\frac{\pi a^2}{4},\qquad \text{ellipse area}=\pi ab\]

Parabola and latus rectum

\[y^2=4ax,\ x=a:\quad A=\frac{8a^2}{3}\]

Key Points

Use symmetry

A circle or ellipse centred at the origin is symmetric in both axes: compute the first-quadrant area and multiply by 4. A parabola y² = 4ax is symmetric about the x-axis: double the upper half.

Common Mistakes

Negative area

\(\int_0^{2\pi}\sin x\,dx=0\), but the area between \(\sin x\) and the axis on \([0,2\pi]\) is \(2+2=4\).

Wrong limits for x = g(y)

When using horizontal strips, the limits must be y-values.

Solved Examples

Area of an ellipse quadrant

Area in the first quadrant enclosed by \(\frac{x^2}{25}+\frac{y^2}{4}=1\).

Practice & Topic Test

Topic Practice

A fresh set of questions from this topic. Change answers freely – solutions appear after you submit.

Start Practice

Topic Test

Timed: up to 10 questions in 15 minutes.

Start Test

Topic Summary

Sketch, choose strips, set limits, integrate. Always give a positive answer in square units.

Ready to practise this topic?

Create a free account to take practice sessions and tests, see detailed explanations and track your progress.

Create Student Account