Application of Integrals — Study Notes

1. Introduction

The application of integrals chapter puts the definite integral to work computing geometric quantities that are difficult or impossible to obtain by elementary geometry. The central idea is that a definite integral accumulates the area between a curve and the x-axis, and by combining and subtracting such areas we can find the areas enclosed by curves, the areas between two curves, and the lengths and volumes of solids of revolution.

The chapter begins with the area under a curve and above the x-axis, distinguishing between the signed area given by the integral and the actual geometric area obtained by taking absolute values. It then develops the area enclosed between a curve and the axes, and finally the area between two curves, which is computed as the integral of the difference of the two functions. These techniques depend directly on a sound grasp of Chapter 7's definite integrals.

This chapter is comparatively short, but it is always examined, typically in the long-answer section, and it offers a reliable opportunity to score full marks with careful diagram drawing and methodical limit setting. The ability to sketch the region correctly and to identify the correct limits of integration is the single most important skill. Beyond the examination, area and volume computation by integration is the basis of all measurement in applied mathematics and physics.

2. Area Under a Curve

Let f be a continuous function on [a, b] with f(x) >= 0. The area enclosed by the curve y = f(x), the x-axis, and the lines x = a and x = b is

$$A = \int_a^b f(x)\, dx$$

If f(x) <= 0 on the interval, the integral is negative, and the area is the absolute value:

$$A = \left| \int_a^b f(x)\, dx \right|$$

For a function that changes sign, split the interval at the zeros and take absolute values in each subinterval.

Example: Area of a Circle Region

The area enclosed by the circle $x^2 + y^2 = a^2$ can be computed by integrating the upper semicircle $y = \sqrt{a^2 - x^2}$ from $-a$ to $a$ and doubling the area in the first quadrant. Since

$$\int_0^a \sqrt{a^2 - x^2}\, dx = \frac{\pi a^2}{4}$$

the total area is $4 \times \frac{\pi a^2}{4} = \pi a^2$.

3. Area Between Two Curves

If f and g are continuous functions on [a, b] with f(x) >= g(x) throughout, then the area between the curves y = f(x) and y = g(x) from x = a to x = b is

$$A = \int_a^b [f(x) - g(x)]\, dx$$

Procedure

1. Sketch both curves and shade the required region.
2. Find the points of intersection of the two curves by solving f(x) = g(x); these give the limits of integration.
3. Determine which function is on top (greater) in the region.
4. Integrate the difference of the upper and lower functions between the limits.

Example

The area enclosed by the parabola $y = x^2$ and the line $y = x$: the points of intersection satisfy $x^2 = x$, giving x = 0 and x = 1. Since the line lies above the parabola on (0, 1):

$$A = \int_0^1 (x - x^2)\, dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$$

4. Areas with Symmetry

Many regions are symmetric about the axes, which allows the area to be computed over a simpler interval and then multiplied. For example, the region bounded by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ has area

$$A = 4\int_0^a \frac{b}{a}\sqrt{a^2 - x^2}\, dx = 4 \cdot \frac{b}{a} \cdot \frac{\pi a^2}{4} = \pi ab$$

5. Volume of a Solid of Revolution

When the region under the curve y = f(x) from a to b is rotated about the x-axis, the volume swept out is

$$V = \pi \int_a^b [f(x)]^2\, dx$$

For example, rotating $y = \sqrt{a^2 - x^2}$ (a semicircle) about the x-axis gives a sphere of volume

$$V = \pi \int_{-a}^{a} (a^2 - x^2)\, dx = \pi \left[a^2x - \frac{x^3}{3}\right]_{-a}^{a} = \frac{4}{3}\pi a^3$$

Quick Revision Tables

Table 1: Area Formulas

Table 2: Procedure for Area Between Curves

Table 3: Volume of Revolution

Mind Map

Important Diagrams (SVG)

Diagram 1: Area Between a Parabola and a Line

Diagram 2: Semicircle and Circle Area via Integration

Common Mistakes

1. Using the signed integral when the curve lies below the x-axis, producing a negative area; take the absolute value.
2. Integrating (lower - upper) instead of (upper - lower), which flips the sign of the area.
3. Taking the intersection points as limits without checking they bound the required region.
4. Forgetting to split the interval where the two curves cross; the upper function changes identity at the crossing point.
5. Drawing the wrong region when the area is enclosed by a curve and the axes; carefully identify the region that satisfies all bounding conditions.
6. In volume problems, forgetting the factor pi and the square: V = pi int [f(x)]^2 dx.
7. Substituting a = b = r into the ellipse area formula incorrectly; the circle is the special case with a = b.
8. Neglecting to shade the region or sketch the curves, losing method marks in the board exam.
9. Using limits from the wrong curve intersections; each pair of boundaries must be solved consistently.
10. Treating area above and below the axis as cancelling; actual area requires adding absolute contributions from each subinterval.

Exam Tips

1. Always begin area problems by sketching the curves and shading the region; examiners award marks for the correct region identification.
2. Find intersection points by equating the functions and solving for x, then verify the bounds enclose the intended region.
3. Test a sample x-value between the limits to decide which function is the upper curve before writing the integrand.
4. Use symmetry aggressively: for circles and ellipses, integrate over a quarter or half and multiply.
5. For the area enclosed between a parabola and a line, the standard limits are the roots of the equality; memorise the setup but re-derive in the exam.
6. In volume questions, square the function first, integrate, then multiply by pi; do not square after integrating.
7. If the region is bounded by curves given in terms of y, consider integrating with respect to y, using horizontal strips and the corresponding left and right functions.
8. Keep answers in terms of pi where the region involves circles or standard areas, unless the question asks for a decimal.

Conclusion

The application of integrals transforms the abstract definite integral into a direct instrument of measurement. By computing areas under curves, between two curves, and volumes of solids of revolution, the chapter demonstrates that the integral is not merely an algebraic operation but the fundamental tool of geometry in calculus. The key discipline is the systematic procedure: sketch the region, determine the limits by solving for intersection points, decide which function bounds the region from above and below, and integrate the correct difference. Symmetry, in particular, reduces many difficult integrals to simple ones. Because the problems are structured and routine once the method is mastered, this chapter is one of the most reliable sources of full marks in the board examination, and it provides the geometric intuition that underpins the study of differential equations and higher-dimensional calculus.