Application of Integrals — Class 12 IMO

Area Under a Curve

The definite integral computes the area between a curve and an axis. For a curve $y=f(x)$ that lies above the $x$-axis between $x=a$ and $x=b$,

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

When the curve dips below the axis

Where $f(x)<0$, the integral returns a negative value. Since area is positive, split the interval at the $x$-intercepts and take the absolute value of each piece:

$$\text{Area}=\int_a^c f\,dx + \left|\int_c^b f\,dx\right|.$$

Area with respect to the $y$-axis

If a region is described more naturally in terms of $x=g(y)$, integrate along $y$:

$$\text{Area}=\int_c^d x\,dy=\int_c^d g(y)\,dy.$$

Always sketch the region first — the picture fixes the limits and tells you whether to integrate in $x$ or $y$.

Example 1: Find the area under $y=x^2$ from $x=0$ to $x=3$.

$\int_0^3 x^2\,dx=\left[\dfrac{x^3}{3}\right]_0^3=\dfrac{27}{3}=9$ square units.

Example 2: Find the area bounded by $y=x^2$, the $x$-axis and $x=2$.

$\int_0^2 x^2\,dx=\left[\dfrac{x^3}{3}\right]_0^2=\dfrac{8}{3}$ square units.

Example 3: Find the area under $y=\sin x$ from $0$ to $\pi$.

$\int_0^{\pi}\sin x\,dx=[-\cos x]_0^{\pi}=-(-1)+1=2$ square units.

Example 4: Find the area of the region bounded by $x=y^2$ and $x=4$ (in the first quadrant).

Integrate along $y$ from $0$ to $2$ (since $x=4\Rightarrow y=2$): area $=\int_0^2 (4-y^2)\,dy=\left[4y-\dfrac{y^3}{3}\right]_0^2=8-\dfrac{8}{3}=\dfrac{16}{3}$ square units.

Quick recap

Area above the $x$-axis $=\int_a^b f(x)\,dx$.
Below the axis the integral is negative; split at intercepts and take absolute values.
Area w.r.t. the $y$-axis $=\int_c^d g(y)\,dy$.
Sketch the region first to set the correct limits.

✓ Quick check

The area enclosed between the curves y = x² and y = x is:

Intersection points: x² = x → x(x − 1) = 0 → x = 0, 1. Area = ∫[0 to 1] (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6 sq unit.

Find the area of the region in the first quadrant enclosed by the circle x² + y² = 4.

Total area = π(2²) = 4π. Area in the first quadrant = 4π / 4 = π sq units.

Area Between Two Curves

The area between two curves $y=f(x)$ (upper) and $y=g(x)$ (lower) from $x=a$ to $x=b$ is the integral of the gap between them:

$$\text{Area}=\int_a^b \big[f(x)-g(x)\big]\,dx,\qquad f(x)\ge g(x)\ \text{on }[a,b].$$

Finding the limits

The limits $a,b$ are usually the $x$-coordinates of the points of intersection, found by solving $f(x)=g(x)$. Decide which curve is on top in the region (test a sample point) so the integrand stays non-negative.

If the curves cross

When the curves swap top/bottom inside the interval, split at the crossing points and integrate each sub-interval with the correct "upper minus lower" order. The total area is the sum of these positive pieces.

Example 1: Find the area between $y=x$ and $y=x^2$ from their intersections.

Intersections: $x=x^2\Rightarrow x=0,1$. On $(0,1)$, $x\ge x^2$. Area $=\int_0^1 (x-x^2)\,dx=\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1=\dfrac12-\dfrac13=\dfrac16$ square units.

Example 2: Find the area between $y=x^2$ and $y=4$.

Intersections: $x^2=4\Rightarrow x=\pm2$. Upper curve is $y=4$. Area $=\int_{-2}^{2}(4-x^2)\,dx=\left[4x-\dfrac{x^3}{3}\right]_{-2}^{2}=\left(8-\dfrac83\right)-\left(-8+\dfrac83\right)=\dfrac{32}{3}$ square units.

Example 3: Find the area between the line $y=2x$ and the parabola $y=x^2$.

$x^2=2x\Rightarrow x=0,2$. On $(0,2)$, $2x\ge x^2$. Area $=\int_0^2(2x-x^2)\,dx=\left[x^2-\dfrac{x^3}{3}\right]_0^2=4-\dfrac83=\dfrac{4}{3}$ square units.

Example 4: Set up the area between $y=\sqrt{x}$ and $y=x$ for $0\le x\le1$.

On $(0,1)$, $\sqrt{x}\ge x$. Area $=\int_0^1(\sqrt{x}-x)\,dx=\left[\dfrac{2}{3}x^{3/2}-\dfrac{x^2}{2}\right]_0^1=\dfrac23-\dfrac12=\dfrac16$ square units.

Quick recap

Area between curves $=\int_a^b(\text{upper}-\text{lower})\,dx$.
Limits are the intersection points: solve $f(x)=g(x)$.
Test a sample point to decide which curve is upper.
If the curves cross, split the interval and add positive pieces.

✓ Quick check

Calculate the area of the region bounded by the parabolas y = 2x² and y = x² + 4.

Intersections: 2x² = x² + 4 → x² = 4 → x=±2. Area = ∫(−2 to 2) (x² + 4 − 2x²) dx = 2∫(0 to 2) (4 − x²) dx = 32/3 sq units.

The area bounded by the curve y = log x, the x-axis and the lines x = 1 and x = e is:

Area = ∫[1 to e] log x dx = [x log x − x]₁ᵉ = (e·1 − e) − (1·0 − 1) = 0 − (−1) = 1 sq unit.

Areas of Standard Regions

Integration recovers familiar areas: a circle x² + y² = a² has area πa², and integration gives the area of segments and parabolic regions.

Exploit symmetry to integrate over one quadrant and multiply.

Example 1: Area enclosed by x² + y² = 4.

π·2² = 4π.

Example 2: Area of a quarter of the circle x² + y² = 9.

(1/4)·9π = 9π/4.

Quick recap

Circle of radius a: area πa².
Use symmetry to shorten the integral.

✓ Quick check

A plot is in the shape of the region between y = √x, y = x. The cost of fencing at ₹50 per unit length of perimeter is approximately (use integration for arc length if needed, but area in sq units is asked here). Actually, if land price is ₹200 per sq unit, the total land cost is:

Area = 1/6 sq unit. Cost = (1/6) × 200 = ₹100/3.

Sachin is creating a logo for the E-learning Hub Yudgam. The design involves the region bounded by y = x² and the horizontal line y = 9. What is the geometric area of this region in square units?

Intersect at x = −3, 3. Area = ∫(−3 to 3) (9 − x²) dx = 2 × [9x − x³/3] from 0 to 3 = 2 × (27 − 9) = 36 sq units.