Application of Integrals

Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XII, Chapter 8.

How do you measure a shape with a curved edge?

A rectangle has length times breadth, a triangle has half base times height, and a circle has $\pi r^2$. But what is the area of the patch of land under a curved road, or the region cut off by a parabola? There is no ready-made formula. The trick of this chapter is to cut the region into very thin vertical or horizontal strips, treat each strip as a tiny rectangle, add them all up, and let the strips get thinner and thinner. That adding-up in the limit is exactly what a definite integral does.

The method in one line: sketch → choose the strip → find where curves meet the axes → split where the sign changes → integrate → take absolute values → add.

The thin strip idea

Take the region under the curve $y = f(x)$, above the $x$-axis, between the lines $x = a$ and $x = b$. Cut a vertical strip of width $dx$ at position $x$. Its height is $y = f(x)$, so its area is approximately

$$dA = y\,dx$$

Adding every strip from $x = a$ to $x = b$ is what the integral sign means:

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

The same idea works sideways. For the region bounded by $x = g(y)$, the $y$-axis and the lines $y = c$ and $y = d$, use a horizontal strip of width $dy$ and length $x$:

$$A = \int_c^d x\,dy = \int_c^d g(y)\,dy$$

Worked example: choosing the strip. Find the area bounded by the parabola $y^2 = 4x$, the $y$-axis and the line $y = 3$.

Common wrong step. Using $x$-limits such as $0$ to $3$ with a vertical strip. The line $y = 3$ is horizontal, so its end of the region is at $y = 3$, not $x = 3$. The limits always belong to the variable you integrate.

When the curve goes below the axis

An integral gives a signed area: areas above the $x$-axis count positive, areas below count negative. A real area can never be negative, so the signs need care.

If the curve lies below the $x$-axis between $x = a$ and $x = b$, then $\int_a^b f(x)\,dx$ is negative, and the area is its absolute value:

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

If the curve crosses the axis inside $[a, b]$, splitting is essential. Let $A_1$ be the integral over the part below the axis (a negative number) and $A_2$ the integral over the part above. Then

$$A = \lvert A_1\rvert + A_2$$

If you integrate across the crossing in one stroke you get $A_1 + A_2$, where the negative piece cancels part of the positive one. That is the single biggest source of wrong answers in this chapter.

Worked example: a line that crosses the axis. Find the area bounded by $y = 3x + 2$, the $x$-axis, and the lines $x = -1$ and $x = 1$.

Common wrong step. Integrating from $-1$ to $1$ in one go. With $F(x) = \dfrac{3x^2}{2} + 2x$, that gives $F(1) - F(-1) = \dfrac{7}{2} + \dfrac{1}{2} = 4$, which is the signed area. The true area is $\dfrac{13}{3}$, larger by the $\dfrac{1}{3}$ that cancelled.

Worked example: a wave. Find the area bounded by $y = \cos x$, the $x$-axis and the lines $x = 0$ and $x = 2\pi$.

Common wrong step. Computing $\int_0^{2\pi} \cos x\,dx = 0$ and reporting an area of zero. The integral is $0$, but the area is $4$.

Pause and predict: what are $\int_0^{2\pi} \sin x\,dx$ and the area between $y = \sin x$ and the $x$-axis from $0$ to $2\pi$?

Check your answer: the integral is $0$ and the area is $4$. The arch above the axis contributes $2$ and the arch below contributes $-2$, so the integral cancels but the area adds $2 + 2$.

Beyond the textbook. For the exam: signed integrals must be converted to areas by taking absolute values piece by piece. In reality: the signed value is not a nuisance. For a velocity-time graph, $\int v\,dt$ is the displacement, in which backward motion counts negative, while the total area (with absolute values) is the distance travelled. Why it matters: the same mathematics answers two different physical questions, and the difference between them is exactly the sign handling in this chapter.

The circle and the ellipse

The circle $x^2 + y^2 = a^2$ is symmetric about both axes, so its four quadrants have equal areas. In the first quadrant $y = \sqrt{a^2 - x^2}$, and $x$ runs from $0$ to $a$. Therefore:

$$A = 4\int_0^a \sqrt{a^2 - x^2}\,dx = 4\left[ \dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\left( \dfrac{x}{a} \right) \right]_0^a = 4\cdot\dfrac{a^2}{2}\cdot\dfrac{\pi}{2} = \pi a^2$$

At $x = a$ the root term is zero and $\sin^{-1} 1 = \dfrac{\pi}{2}$, and at $x = 0$ both terms are zero. Notice the sanity check built into the calculation: $\int_0^a \sqrt{a^2 - x^2}\,dx = \dfrac{\pi a^2}{4}$ is a quarter of a circle, exactly what geometry says.

The ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ is the same story. In the first quadrant $y = \dfrac{b}{a}\sqrt{a^2 - x^2}$, which is the circle's height scaled by the constant $\dfrac{b}{a}$, so its area is the circle's area scaled by the same constant:

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

Here $a$ and $b$ are the semi-axes, the distances from the centre to the ends on each axis.

Worked example: an ellipse from its equation. Find the area enclosed by $4x^2 + 9y^2 = 36$.

Common wrong step. Reading $a = 9$ and $b = 4$ straight from the denominators. They are $a^2$ and $b^2$, so take the square roots first. Writing the area as $\pi a^2 b^2$ is the same slip.

Pause and predict: what does the ellipse formula give when $a = b$?

Check your answer: $\pi a \cdot a = \pi a^2$, the circle. An ellipse with equal semi-axes is a circle, so the formulas agree.

Area between two curves

When a region sits between the graphs of $y = f(x)$ above and $y = g(x)$ below, with $f(x) \ge g(x)$ on $[a, b]$, a vertical strip has height $f(x) - g(x)$:

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

Remember it as top minus bottom. For horizontal strips it becomes right minus left, $\int_c^d \left[ x_{\text{right}} - x_{\text{left}} \right] dy$. The limits are the $x$-values where the curves meet, so solve $f(x) = g(x)$ first. If the curves cross more than once, split the interval at every crossing and apply top minus bottom on each part with the right curve on top.

Worked example: a line and a parabola. Find the area enclosed between $y = x$ and $y = x^2$.

Common wrong step. Taking bottom minus top, which gives $-\dfrac{1}{6}$. An area cannot be negative, so the order must be top minus bottom. The sketch decides which is which, so never skip it.

Using symmetry to save work. If the region is symmetric about the $y$-axis (an even function on $[-p, p]$), the area is twice the right half. For the parabola $y = ax^2$ with $a > 0$ between $x = -p$ and $x = p$, the area under it is $2\int_0^p ax^2\,dx = \dfrac{2}{3}a p^3$. For an odd function the signed integral over $[-p, p]$ is zero, but the area is still $2\int_0^p \lvert f(x)\rvert\,dx$, because the two halves are mirror images with opposite signs.

Beyond the textbook. For the exam: areas under curves are computed from given equations, with answers in square units. In reality: the area under a curve is the working tool of probability. When a quantity follows a smooth density curve, such as the bell curve for heights or marks, the probability of a value lying between $a$ and $b$ is the area under the curve between them, and the total area under the whole curve is $1$. Why it matters: this is why statistics tables are really tables of integrals that nobody can write in a simple closed form.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Integrals supplies the second fundamental theorem and the $\int\sqrt{a^2 - x^2}\,dx$ formula used for the circle and the ellipse.
• Next: Differential Equations puts integration to work in solving equations that involve rates of change.
• Earlier ideas reused: Conic Sections gives the equations of the circle, ellipse and parabola, and Straight Lines gives lines like $y = 3x + 2$.
• Same family of ideas: Application of Derivatives uses the same curve-sketching habits for rates and extremes.