Area under a curve
Applications of Integrals · Area Under Simple Curves\[A=\int_a^b y\,dx\quad\text{or}\quad A=\int_c^d x\,dy\]
Chapter Overview
A definite integral of a non-negative function is the area under its graph. In this chapter you will use integration to find areas of regions bounded by simple curves – straight lines, parabolas, circles and ellipses in standard form – and the coordinate axes or vertical/horizontal lines. You will see that the familiar formulas \(\pi r^2\) and \(\pi ab\) come out of integration.
Board focus: One long-answer (5-mark) area question is common: sketch the region, set up the integral, evaluate.
Topics
Key Concepts
- Area between \(y=f(x)\ (\ge0)\), the \(x\)-axis and \(x=a,\ x=b\): \(\int_a^bf(x)\,dx\).
- Area between \(x=g(y)\ (\ge0)\), the \(y\)-axis and \(y=c,\ y=d\): \(\int_c^dg(y)\,dy\).
- If the curve dips below the axis, take the absolute value of each part separately.
- Use symmetry: circle and ellipse areas are 4 × (first-quadrant area).
Formulas
Circle and ellipse
Applications of Integrals · Area Under Simple Curves\[\int_0^a\sqrt{a^2-x^2}\,dx=\frac{\pi a^2}{4},\qquad \text{ellipse area}=\pi ab\]
Parabola and latus rectum
Applications of Integrals · Area Under Simple Curves\[y^2=4ax,\ x=a:\quad A=\frac{8a^2}{3}\]
Key Points
Use symmetry
Applications of Integrals · Area Under Simple CurvesA 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
Applications of Integrals · Area Under Simple Curves\(\int_0^{2\pi}\sin x\,dx=0\), but the area between \(\sin x\) and the axis on \([0,2\pi]\) is \(2+2=4\).
Wrong limits for x = g(y)
Applications of Integrals · Area Under Simple CurvesWhen using horizontal strips, the limits must be y-values.
Solved Examples
Area of an ellipse quadrant
Applications of Integrals · Area Under Simple CurvesArea in the first quadrant enclosed by \(\frac{x^2}{25}+\frac{y^2}{4}=1\).
\(y=\frac25\sqrt{25-x^2}\), so the area is \(\frac25\int_0^5\sqrt{25-x^2}\,dx=\frac25\cdot\frac{25\pi}{4}=\frac{5\pi}{2}\).
Practice & Tests
Chapter Practice
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