Area Between Two Curves in Calculus

The area between two curves in calculus refers to the region enclosed between two functions over a given interval. It is calculated using definite integrals by subtracting the lower function from the upper function. This concept helps in solving problems involving enclosed areas in mathematics, physics, and engineering. The formula applies to both horizontal and vertical distances and can also be extended to polar curves. Mastering this topic is essential for Class 12 and competitive exams like JEE.

1.0Area Between Two Curves Definition

The area between two curves refers to the region enclosed between the graphs of two functions over a given interval. It represents the net difference of their values across a certain range of x or y.

2.0Formula for the Area Between Two Curves

If two continuous functions f(x) and g(x) satisfy $f(x)\ge g(x)$ on the interval [a, b], then the area between the curves with respect to x is: $Area={\int }_a^b|f(x)-g(x)|dx$

Formula for the Area Between Two Curves with Respect to y:

If functions are expressed as x = f(y) and x = g(y): $Area={\int }_c^d|f(y)-g(y)|dy$

3.0How to Find the Area Under a Curve That is Between Two Values?

1. Find the points of intersection by solving f(x) = g(x).
2. Set correct limits of integration based on the intersection points.
3. Identify which function is on top (upper curve) and which is below (lower curve).
4. Use the area between two curves formula with correct limits.
5. Solve the definite integral to get the area.

4.0Area Between Two Compound Curves

In cases involving compound curves (multiple intersections or changing upper/lower curves), split the integration into sections where one curve stays above the other and apply the area formula separately for each section.

5.0Area Between Two Polar Curves

For curves defined in polar form $r=f(\theta )$ and $r=g(\theta )$, the area between two polar curves is:$Area=\frac{1}{2}{\int }_{{\theta }_1}^{{\theta }_2}|[f(\theta ){]}^2-[g(\theta ){]}^2|d\theta$

6.0Solved Examples on Area Between Two Curves

Example 1: Find the area bounded between the curves $y=x^2$ and y = 2x.

Solution:

1. Find intersection points: $x^2=2x\to x=0orx=2.$
2. Integrate: ${\int }_0^2(2x-x^2)dx=[x^2-\frac{x^3}{3}{]}_0^2$$=4-\frac{8}{3}=\frac{4}{3}$

Answer: $\frac{4}{3}$square units.

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

Solution:

1. Solve: $y^2=4y$$\to$y = 0, 4.
2. Integrate with respect to y:

${\int }_0^4(4y-y^2)dy=[2y^2-\frac{y^3}{3}{]}_0^4$ $=2(16)-\frac{64}{3}=\frac{64}{3}$

Answer: $\frac{64}{3}$ square units.

Example 3: Find the area bounded by the curves $y^2=4x$ and y = 2x.

Solution:

1. Solve intersection points:

$y^2=4x\to x=\frac{y^2}{4}$

$y=2x\to x=\frac{y}{2}$

Set them equal:

$\frac{y^2}{4}=\frac{y}{2}$

$\Rightarrow y^2=2y$

$\Rightarrow y(y-2)=0$

$\Rightarrow y=0,2$

2. Area w.r.t y:

${\int }_0^2\left(\frac{y}{2}-\frac{y^2}{4}\right)dy$

3. Solve:

${\int }_0^2(y2-y24)dy=[y24-y312{]}_0^2$

${\int }_0^2\left(\frac{y}{2}-\frac{y^2}{4}\right)dy={\left[\frac{y^2}{4}-\frac{y^3}{12}\right]}_0^2$

$=\left(\frac{4}{4}-\frac{8}{12}\right)=1-\frac{2}{3}=\frac{1}{3}$

Answer: $\frac{1}{3}$ square units.

Example 4: Find the area enclosed between the polar curves $r=cos\theta$ and $r=sin\theta$.

Solution:

1. Find points of intersection: $cos\theta =sin\theta \Rightarrow \theta =\frac{\pi }{4}$
2. Area between the curves from 0 to $\frac{\pi }{4}$: $\frac{1}{2}{\int }_0^{\frac{\pi }{4}}(si{n}^2\theta -co{s}^2\theta )d\theta$
3. Solve: $\frac{1}{2}{\int }_0^{\frac{\pi }{4}}(-cos2\theta )d\theta$
4. Integrate:

$=-\frac{1}{4}sin2\theta {|}_0^{\frac{\pi }{4}}=-\frac{1}{4}(sin(\frac{\pi }{2}-sin0)$

$=-\frac{1}{4}(1-0)=-\frac{1}{4}$

Since area is positive:

Answer: $\frac{1}{4}$square units.

Example 5: Find the area bounded by y = |x| and y = 4 - |x|.

Solution:

1. Solve intersection points:

$x|=4-|x|$

$\Rightarrow |x|=2$

$\Rightarrow =\pm 2$

2. Area symmetric about y-axis, compute for $0\le x\le 2$ and double:

$2\times {\int }_0^2\left[(4-x)-x\right]dx$

$=2\times {\int }_0^2(4-2x)dx$

$=2\times {\left[4x-x^2\right]}_0^2$

$=2\times (8-4)=8$

Answer: 8 square units.

Example 6: Find the area between the curves $y=x^3$ and y = x in the interval [-1, 1].

Solution:

1. Solve intersection points:

$x^3=x$

$\Rightarrow x(x^2-1)=0$

$\Rightarrow =-1,0,1$

2. Split into intervals and integrate:

$\text{Area}={\int }_{-1}^0[x-x^3]dx+{\int }_0^1[x-x^3]dx$

Because the integrand is odd over symmetric limits,

$\text{Area}=2\times {\int }_0^1(x-x^3)dx$

$=2\times {\left[\frac{x^2}{2}-\frac{x^4}{4}\right]}_0^1=2\times \left(\frac{1}{2}-\frac{1}{4}\right)=2\times \frac{1}{4}=\frac{1}{2}$ 

Answer: $\frac{1}{2}$ square units.

Example 7: Find the area enclosed between the polar curves $r=2sin\theta$ and r = 1 for $0\le \theta \le \frac{\pi }{2}$.

Solution:

1. Use the formula:$\frac{1}{2}{\int }_{{\theta }_1}^{{\theta }_2}|[f(\theta ){]}^2-g(\theta ){]}^2|d\theta$
2. Apply:$\frac{1}{2}{\int }_0^{\frac{\pi }{2}}|(2sin\theta {)}^2-1^2|d\theta$
3. Simplify:$\frac{1}{2}{\int }_0^{\frac{\pi }{2}}|4si{n}^2\theta -1|d\theta$

Solve this integral step by step using trigonometric identities.

Example 8: Find the area enclosed between the curves y = sin x and y = cos x in $[0,\frac{\pi }{4}]$.

Solution:

1. Identify upper curve: In $[0,\frac{\pi }{4}],cosx\ge sinx$
2. Setup integral: ${\int }_0^{\frac{\pi }{4}}(cosx-sinx)dx$
3. Solve:

$={\left[\sin x+\cos x\right]}_0^{\frac{\pi }{4}}$

$=\sin \frac{\pi }{4}+\cos \frac{\pi }{4}-(\sin 0+\cos 0)$

$=\sqrt{2}-1$

Answer: $\sqrt{2}-1$ square units.

Example 9: Find the area bounded by $y=x^2$ and $y=x^3$.

Solution:

1. Solve $x^2=x^3\Rightarrow x^2(1-x)=0\Rightarrow x=0,1$
2. Integrate:

${\int }_0^1\left(x^2-x^3\right)dx$

$={\left[\frac{x^3}{3}-\frac{x^4}{4}\right]}_0^1$

$=\frac{1}{3}-\frac{1}{4}=\frac{1}{12}$

Answer: $\frac{1}{12}$ square units.

7.0Practice Questions based on Area Between Two Curves 

1. Find the area bounded between y = sin x and y = cos x in $[0,\frac{\pi }{4}]$.
2. Find the area between $y=e^x$ and y = ln x over their common domain.
3. Calculate the area between $r=2cos\theta$ and r = 1 in polar coordinates.

8.0Applications of Area Between Two Curves

• Physics (work done by forces)
• Economics (surplus calculation)
• Probability & Statistics (distribution functions)
• Engineering (cross-sectional designs)
• Geometry (irregular shapes)

9.0Sample Questions on Area Between Two Curves

1. What is the formula for the area between two curves?

Ans: If f(x) is the upper curve and g(x) is the lower curve in the interval [a, b], then: $Area={\int }_0^b[f(x)-g(x)]dx$

2. Can we calculate the area between two curves with respect to y?

Ans: Yes. If the curves are expressed as x = f(y) and x = g(y), then: $Area={\int }_c^d[f(y)-g(y)]dy$

3. What is the formula for the area between two curves in polar coordinates?

Ans: For polar curves $r=f(\theta )$ and $r=g(\theta )$: $Area=\frac{1}{2}{\int }_{{\theta }_1}^{{\theta }_2}|[f(\theta ){]}^2-g(\theta ){]}^2|d\theta$