- Chapter 8
- Calculus
- 20 questions
Application of Integrals
Area under simple curves, and areas bounded by lines, circles, parabolas and ellipses in standard form.
A definite integral measures accumulated area. Once that is accepted, finding the region enclosed between curves becomes a matter of care rather than of new theory: identify the region, find where the boundaries meet, decide which curve is on top, and integrate the difference.
The single most common source of lost marks here is not the integration at all — it is a sketch that was never drawn, and a region that was therefore misidentified.
What you should be able to do
- Find the area under a simple curve between given limits.
- Find the area bounded by a curve and a line, or between two curves.
- Handle regions bounded by circles, parabolas and ellipses given in standard form.
- Use symmetry to reduce the work, and interpret areas that fall below the axis correctly.
Topics
2 topics
- Area Under Simple Curves
Area bounded by a curve, the axis and two ordinates, including regions below the axis.
- Area Bounded by Standard Curves
Areas involving lines, circles, parabolas and ellipses in standard form.
Formulas
7 items · All formulas
Area under a curve (with respect to $x$)
A = \int_{a}^{b} y\,dx = \int_{a}^{b} f(x)\,dx- Valid when
- Valid where f(x) \geq 0; otherwise take the absolute value piecewise.
Area under a curve (with respect to $y$)
A = \int_{c}^{d} x\,dyUse horizontal strips when they cross the region without splitting.
Area below the axis
A = \left|\int_{a}^{b} f(x)\,dx\right|- Valid when
- When f(x) \leq 0 throughout [a,b].
Area between two curves
A = \int_{a}^{b}\left(y_{\text{upper}} - y_{\text{lower}}\right)dx- Valid when
- a and b are the x-coordinates of the intersection points.
Area of a circle by integration
A = 4\int_{0}^{a}\sqrt{a^{2} - x^{2}}\,dx = \pi a^{2}One quadrant computed and multiplied by four, using symmetry.
Area of an ellipse by integration
A = 4\int_{0}^{a}\frac{b}{a}\sqrt{a^{2}-x^{2}}\,dx = \pi a b- Valid when
- For \dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1.
Standard integral needed for circles
\int \sqrt{a^{2}-x^{2}}\,dx = \frac{x}{2}\sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2}\sin^{-1}\!\left(\frac{x}{a}\right) + C
Key points
6 items · All key points
Always sketch the region before writing any integral.
The sketch fixes the limits, identifies which curve is on top, and reveals any crossing of the axis.
If the curve crosses the x-axis inside the interval, split the integral there and add the absolute values.
Integrating straight through lets the positive and negative parts cancel, understating the area.
Area is always positive. A negative result means the region lies below the axis — take the modulus.
For area between curves, subtract lower from upper — and check which is which on the interval concerned.
Exploit symmetry for circles and ellipses: compute one quadrant and multiply.
Find the limits by solving the curves simultaneously, not by reading them off the question.
Common mistakes
5 items · All common mistakes
- Mistake
Reporting a negative value as the area of a region below the axis.
InsteadTake the absolute value. Area is a magnitude and cannot be negative.
WhyThe definite integral computes signed area, and the sign is carried through to the final line unthinkingly.
- Mistake
Integrating across a point where the curve crosses the axis without splitting.
InsteadSplit at the crossing and add \left|\int\right| for each piece.
WhyWithout a sketch, the crossing goes unnoticed.
- Mistake
Subtracting the wrong way round when finding the area between two curves.
InsteadTest a point inside the interval to see which curve is genuinely on top there.
- Mistake
Guessing the limits instead of solving the two equations simultaneously.
InsteadSet the curves equal and solve; the roots are the x-coordinates of the limits.
- Mistake
Using \int y\,dx when the region is bounded on the left and right by curves.
InsteadSwitch to \int x\,dy with horizontal strips, which crosses such a region cleanly.