Application Of Integrals — Complete Guide with Solved Examples

What Are We Actually Doing Here?

Integration gives us areas. That’s the core idea — and once you see it clearly, the entire chapter becomes predictable.

When we integrate a function $f(x)$ from $a$ to $b$ , we’re computing the signed area between the curve and the x-axis. “Signed” means areas below the x-axis come out negative. In this chapter, we use that machinery to find actual geometric areas — and occasionally volumes (though volumes of revolution is a separate topic).

The two main tasks in this chapter are:

Area under a curve — between a curve and the x-axis (or y-axis)
Area between two curves — the region sandwiched between $f(x)$ and $g(x)$

That’s it. The chapter looks longer because of the different cases: which curve is on top, what happens when they intersect, whether we integrate along x or y. But the underlying idea is always subtraction of integrals.

JEE Main typically gives one question from this chapter — worth 4 marks. CBSE gives a 5-mark question (sometimes 3-mark) in Section E of the board paper. This makes it a high-value, low-effort scoring topic if you’ve practiced the curve sketching.

Key Terms and Definitions

Definite Integral as Area: For a curve $y = f(x)$ that lies above the x-axis on $[a, b]$ :

\text{Area} = \int_a^b f(x)\, dx

Signed Area: If the curve dips below the x-axis on part of the interval, $\int_a^b f(x)\, dx$ gives a value where below-axis portions are subtracted. For actual area (always positive), we take the absolute value of each portion.

Area Between Two Curves: When $f(x) \geq g(x)$ on $[a, b]$ :

\text{Area} = \int_a^b [f(x) - g(x)]\, dx

Here $f(x)$ is the upper curve and $g(x)$ is the lower curve. The order matters.

Points of Intersection: To find limits of integration when curves bound a region, solve $f(x) = g(x)$ . These x-values become your limits $a$ and $b$ .

Integrating Along y-axis: Sometimes integrating w.r.t. $y$ is easier — especially for parabolas opening left/right. The formula becomes:

\text{Area} = \int_c^d [x_{\text{right}} - x_{\text{left}}]\, dy

Core Methods with Worked Examples

Method 1: Area Under a Single Curve

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

If $f(x) \geq 0$ throughout $[a, b]$ : $A = \int_a^b f(x)\, dx$

If the curve crosses x-axis at $c \in (a, b)$ : split as $\int_a^c f(x)\, dx + \int_c^b f(x)\, dx$ (take absolute values of each part)

Worked Example (CBSE level): Find the area bounded by $y = x^2$ , the x-axis, and the lines $x = 1$ , $x = 3$ .

The parabola $y = x^2$ is always above the x-axis. So:

A = \int_1^3 x^2\, dx = \left[\frac{x^3}{3}\right]_1^3 = \frac{27}{3} - \frac{1}{3} = \frac{26}{3} \text{ sq. units}

Method 2: Area Between Two Curves

The key step before integrating is always sketching — identify which curve is on top.

Worked Example (JEE Main level): Find the area enclosed by $y = x^2$ and $y = x$ .

Step 1 — Find intersection points. Set $x^2 = x \Rightarrow x(x-1) = 0 \Rightarrow x = 0, 1$ .

Step 2 — Determine which is on top. At $x = 0.5$ : $y = x$ gives $0.5$ , $y = x^2$ gives $0.25$ . So $y = x$ is the upper curve.

Step 3 — Integrate.

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} \text{ sq. units}

The area enclosed by $y = x^n$ and $y = x^m$ (where $m < n$ for $x \in [0,1]$ ) is always $\frac{1}{m+1} - \frac{1}{n+1}$ . Many JEE options are designed around this pattern — recognize it fast.

Method 3: Area Using y as the Variable

Some regions are ugly to integrate w.r.t. $x$ but clean w.r.t. $y$ . A parabola like $x = y^2$ is a classic case.

Worked Example: Find the area bounded by $x = y^2$ and $x = 4$ .

The parabola $x = y^2$ opens right. At $x = 4$ : $y = \pm 2$ . The region spans $y \in [-2, 2]$ .

For each $y$ , the right boundary is $x = 4$ and the left boundary is $x = y^2$ .

A = \int_{-2}^{2} (4 - y^2)\, dy = 2\int_0^2 (4 - y^2)\, dy

(Used symmetry about x-axis)

= 2\left[4y - \frac{y^3}{3}\right]_0^2 = 2\left(8 - \frac{8}{3}\right) = 2 \cdot \frac{16}{3} = \frac{32}{3} \text{ sq. units}

Solved Examples: Easy to Hard

Easy (CBSE 3-mark)

Q: Find the area under $y = \sin x$ from $x = 0$ to $x = \pi$ .

A = \int_0^\pi \sin x\, dx = [-\cos x]_0^\pi = (-\cos\pi) - (-\cos 0) = 1 + 1 = 2 \text{ sq. units}

CBSE loves $\sin x$ and $\cos x$ area questions. The area under one full arch of $\sin x$ (from 0 to $\pi$ ) is exactly 2 — memorize this; it saves calculation time in the 3-hour paper.

Medium (CBSE 5-mark / JEE Main)

Q: Find the area of the region bounded by $y^2 = 4x$ and $x = 3$ .

The parabola $y^2 = 4x$ opens right, vertex at origin. At $x = 3$ : $y = \pm 2\sqrt{3}$ .

Using symmetry:

A = 2\int_0^3 \sqrt{4x}\, dx = 2\int_0^3 2\sqrt{x}\, dx = 4\int_0^3 x^{1/2}\, dx

= 4 \cdot \left[\frac{2x^{3/2}}{3}\right]_0^3 = \frac{8}{3} \cdot 3^{3/2} = \frac{8}{3} \cdot 3\sqrt{3} = 8\sqrt{3} \text{ sq. units}

Hard (JEE Main 2024 pattern)

Q: Find the area enclosed between the circle $x^2 + y^2 = 4$ and the line $x + y = 2$ in the first quadrant.

Setup: The circle has radius 2. The line $x + y = 2$ intersects the circle where $x^2 + (2-x)^2 = 4$ , giving $x = 0$ or $x = 2$ . Points of intersection: $(0, 2)$ and $(2, 0)$ .

We want the area between the arc and the chord in the first quadrant — the smaller region below the arc, above the line.

A = \int_0^2 \left(\sqrt{4 - x^2} - (2-x)\right) dx

= \int_0^2 \sqrt{4-x^2}\, dx - \int_0^2 (2-x)\, dx

First integral — area under the circle arc from 0 to 2. Using the formula $\int_0^a \sqrt{r^2 - x^2}\, dx = \frac{\pi r^2}{4}$ (quarter circle area) with $r = 2$ :

\int_0^2 \sqrt{4-x^2}\, dx = \frac{1}{4}\pi(4) = \pi

Second integral:

\int_0^2 (2-x)\, dx = \left[2x - \frac{x^2}{2}\right]_0^2 = 4 - 2 = 2

A = \pi - 2 \text{ sq. units}

The combination of a circle arc and a chord is JEE’s favourite setup. Always split the integral: (area under curve) minus (area under line). Memorize $\int_0^r \sqrt{r^2 - x^2}\, dx = \frac{\pi r^2}{4}$ — it saves 3 minutes.

Exam-Specific Tips

CBSE Board (Class 12)

The board paper typically has one area question in Section E (5 marks) and sometimes a 3-marker in Section B. The 5-mark question almost always involves:

A parabola and a line, OR
A parabola and another parabola

CBSE marking scheme awards 1 mark for setting up limits, 2 marks for correct integration, 1 mark for substitution, 1 mark for final answer. Even if your final answer is wrong, show the setup clearly — you get partial credit.

JEE Main

One question, 4 marks. The area between a circle and a line ( $\pi - 2$ type) appeared in JEE Main January 2024. Parabola-line combinations with irrational answers like $\frac{4}{3}$ or $\frac{32}{3}$ are standard.

The fastest students sketch first, integrate second. Spending 30 seconds on a rough sketch prevents sign errors and ensures you pick the correct upper/lower curve.

JEE Main often sets options as $\frac{1}{6}$ , $\frac{1}{3}$ , $\frac{2}{3}$ , $\frac{5}{6}$ for the parabola-line area. If your answer isn’t matching, check whether you flipped the upper and lower curves — this is the most common calculation error in this chapter.

SAT (Math)

SAT doesn’t test definite integration directly, but area under a curve appears in data analysis questions using Riemann sum approximations. Understanding that integration gives accumulated area helps with rate-of-change problems in SAT Section 3.

Common Mistakes to Avoid

Mistake 1 — Ignoring negative signed areas. If the curve goes below the x-axis, $\int_a^b f(x)\, dx$ will subtract that portion. Always sketch first. If the curve crosses the axis at $x = c$ , split into $\int_a^c |f(x)|\, dx + \int_c^b |f(x)|\, dx$ .

Mistake 2 — Swapping upper and lower curves. In $\int_a^b [f(x) - g(x)]\, dx$ , if you put the lower curve first, you get a negative answer. Check at any test point inside $[a, b]$ which function gives a larger value.

Mistake 3 — Wrong limits from intersection points. When finding intersections, solve carefully. For $y = x^2$ and $y = 2x$ : setting equal gives $x^2 = 2x \Rightarrow x = 0$ or $x = 2$ . Don’t just write $x = 2$ and miss $x = 0$ .

Mistake 4 — Forgetting the factor of 2 for symmetric regions. When a region is symmetric about the x-axis (like the parabola $y^2 = 4x$ ), many students integrate from $-r$ to $r$ and forget to check their sign conventions. The cleanest approach: integrate the upper half from 0 to the limit, then double it.

Mistake 5 — Using the wrong variable. If integrating w.r.t. $y$ , your limits must be $y$ -values and your integrand must be in terms of $y$ . A common slip: using $x = y^2$ but substituting $x$ -limits like $x = 0$ to $x = 4$ instead of $y = -2$ to $y = 2$ .

Practice Questions

Q1 (Easy): Find the area bounded by $y = 4 - x^2$ and the x-axis.

Set $4 - x^2 = 0 \Rightarrow x = \pm 2$ . The parabola is above the x-axis between $x = -2$ and $x = 2$ .

A = \int_{-2}^{2} (4-x^2)\, dx = 2\int_0^2 (4-x^2)\, dx = 2\left[4x - \frac{x^3}{3}\right]_0^2 = 2\left(8 - \frac{8}{3}\right) = \frac{32}{3} \text{ sq. units}

Q2 (Easy): Find the area enclosed by $y = x^3$ , x-axis, $x = -1$ , $x = 1$ .

The curve $y = x^3$ is negative for $x \in [-1, 0)$ and positive for $x \in (0, 1]$ .

A = \int_{-1}^{0} |x^3|\, dx + \int_0^1 x^3\, dx = \int_{-1}^{0} (-x^3)\, dx + \int_0^1 x^3\, dx

= \left[-\frac{x^4}{4}\right]_{-1}^0 + \left[\frac{x^4}{4}\right]_0^1 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \text{ sq. units}

Q3 (Medium): Find the area of region enclosed by $y = x^2$ and $y = 2x$ .

Intersection: $x^2 = 2x \Rightarrow x = 0, 2$ . At $x = 1$ : $2x = 2 > 1 = x^2$ , so $y = 2x$ is on top.

A = \int_0^2 (2x - x^2)\, dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3} \text{ sq. units}

Q4 (Medium): Find the area between $y = x^2 - 4x$ and the x-axis.

Set $x^2 - 4x = 0 \Rightarrow x(x-4) = 0 \Rightarrow x = 0, 4$ . The curve opens upward (positive leading coefficient) but the vertex is below x-axis — check: $f(2) = 4 - 8 = -4 < 0$ . So the curve is below x-axis on $(0, 4)$ .

A = \left|\int_0^4 (x^2 - 4x)\, dx\right| = \left|\left[\frac{x^3}{3} - 2x^2\right]_0^4\right| = \left|\frac{64}{3} - 32\right| = \left|\frac{64 - 96}{3}\right| = \frac{32}{3} \text{ sq. units}

Q5 (Medium): Using integration, find the area of the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ .

By symmetry, total area = $4 \times$ (area in first quadrant).

In first quadrant: $y = \frac{2}{3}\sqrt{9 - x^2}$ , $x$ from 0 to 3.

A = 4\int_0^3 \frac{2}{3}\sqrt{9-x^2}\, dx = \frac{8}{3}\int_0^3 \sqrt{9-x^2}\, dx = \frac{8}{3} \cdot \frac{\pi(3)^2}{4} = \frac{8}{3} \cdot \frac{9\pi}{4} = 6\pi \text{ sq. units}

General formula: Area of ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\pi ab$ .

Q6 (Hard): Find the area bounded by the parabola $y = x^2$ and the lines $y = x$ and $y = 2x$ for $x \geq 0$ .

Find intersections: $x^2 = x \Rightarrow x = 0, 1$ and $x^2 = 2x \Rightarrow x = 0, 2$ .

For $x \in [0,1]$ : ordering is $x^2 \leq x \leq 2x$ (check at $x = 0.5$ : $0.25, 0.5, 1$ ). For $x \in [1,2]$ : at $x = 1.5$ : $2.25, 1.5, 3$ — so $x \leq x^2$ for $x > 1$ , meaning $y = x$ is now below $y = x^2$ .

The enclosed region between all three is the area between $y = x$ and $y = x^2$ from 0 to 1, plus the area between $y = 2x$ and $y = x^2$ from 0 to 2, minus overlap. A cleaner reading: region bounded by $y = 2x$ (top), $y = x^2$ (bottom) minus region bounded by $y = x$ (top), $y = x^2$ (bottom).

A = \int_0^2 (2x - x^2)\, dx - \int_0^1 (x - x^2)\, dx = \frac{4}{3} - \frac{1}{6} = \frac{7}{6} \text{ sq. units}

Q7 (Hard — JEE Main type): Find the area enclosed between $y^2 = x$ and $x^2 = y$ .

Rewrite as $y = \sqrt{x}$ and $y = x^2$ . Intersection: $\sqrt{x} = x^2 \Rightarrow x = 0, 1$ .

At $x = 0.25$ : $\sqrt{0.25} = 0.5$ , $(0.25)^2 = 0.0625$ . So $y = \sqrt{x}$ is on top.

A = \int_0^1 (\sqrt{x} - x^2)\, dx = \left[\frac{2x^{3/2}}{3} - \frac{x^3}{3}\right]_0^1 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \text{ sq. units}

Q8 (Hard): Find the area of the region in the first quadrant enclosed by $y = \sin x$ , $y = \cos x$ , and the y-axis.

In $[0, \pi/2]$ : $\sin x$ and $\cos x$ intersect at $x = \pi/4$ . For $x \in [0, \pi/4]$ : $\cos x \geq \sin x$ . For $x \in [\pi/4, \pi/2]$ : $\sin x \geq \cos x$ . But the region enclosed by both curves and the y-axis is from $x = 0$ to $x = \pi/4$ (where the y-axis, $y = \sin x$ , and $y = \cos x$ form a closed region).

A = \int_0^{\pi/4} (\cos x - \sin x)\, dx = [\sin x + \cos x]_0^{\pi/4}

= \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) - (0 + 1) = \sqrt{2} - 1 \text{ sq. units}

FAQs

What is the difference between “area under the curve” and “definite integral”?

They’re the same when the curve is entirely above the x-axis. When the curve dips below, the definite integral gives a signed value (negative portions subtract), but the actual geometric area is always positive. For area, always take $\int |f(x)|\, dx$ or split at the x-axis crossings.

How do I know when to integrate with respect to y instead of x?

Integrate w.r.t. $y$ when the curve is naturally expressed as $x = g(y)$ — like $x = y^2$ , or when the region has a complicated shape horizontally but is clean vertically. Also use it when integrating w.r.t. $x$ would require splitting the integral at multiple points.

What formula is used for the area of a circle using integration?

For a circle $x^2 + y^2 = r^2$ : use symmetry and compute $4\int_0^r \sqrt{r^2 - x^2}\, dx$ . The key result is $\int_0^r \sqrt{r^2 - x^2}\, dx = \frac{\pi r^2}{4}$ (quarter-circle area). This gives total area $= \pi r^2$ .

Can the area between two curves ever come out negative?

If you calculate $\int_a^b [f(x) - g(x)]\, dx$ and get a negative number, it means you had $f(x)$ and $g(x)$ switched — $g$ was actually on top. Area is always positive; if your answer is negative, flip the order of subtraction.

How do I find the area when curves intersect at more than two points?

Split the interval at each intersection point and compute the integral separately for each sub-interval (checking which curve is on top in each region). Add all the absolute values.

What is the area enclosed by $y = x^n$ and $y = x^m$ for $0 \leq x \leq 1$ ?

For $m < n$ (so $x^m > x^n$ on $(0,1)$ ):

A = \int_0^1 (x^m - x^n)\, dx = \frac{1}{m+1} - \frac{1}{n+1}

This formula covers a huge number of JEE and CBSE variants. Worth memorizing.

How many marks does this chapter carry in JEE Main?

Typically 4 marks (one question). The chapter has appeared consistently — once per paper in recent years. Expected pattern: area bounded by a conic (parabola or circle) and a line, with the answer involving $\pi$ or a simple fraction. Preparation time of 3-4 hours gives good ROI for this chapter.

Do I need to prove integration formulas or just apply them?

For CBSE boards, application is sufficient — you won’t be asked to derive the area formula. For JEE, same story. However, knowing why $\int_0^r \sqrt{r^2 - x^2}\, dx = \frac{\pi r^2}{4}$ (it’s the quarter-circle area by geometry) helps you apply it correctly under pressure.