Area Under Curve

The concept of the area under a curve is a fundamental topic in calculus and mathematical analysis. It involves calculating the region bounded by a curve and the x-axis within a specific interval. This measurement is essential for solving problems related to integrals, physics, and statistical applications. In this blog, we will explore the concept of the area under a curve, its applications, and provide a few practice problems, especially those relevant to JEE Mains questions.

1.0Area Under a Curve in Calculus

The area under a curve in calculus is determined using definite integrals. If f(x) is a continuous function defined over an interval [a, b], the area under the curve from x = a to x = b is represented as:

$\text{ Area }={\int }_a^{b}f(x)dx$

This formula helps compute the exact area under the curve of f(x), considering $f(x)\ge 0$ over the interval [a, b]. If f(x) takes negative values, the integral measures the net area, considering regions below the x-axis as negative.

2.0Derivation and Formula

The area under the curve formula can be visualized geometrically. Consider the curve f(x) divided into n small rectangles with equal width $\Delta x$. The area of each rectangle is approximately $f\left(x_i\right)\Delta x$ . By summing up these areas and taking the limit as $\Delta x\to 0$ , we obtain:

$\text{ Area }={\lim }_{\Delta x\to 0}{\sum }_{i=1}^{n}f\left(x_i\right)\Delta x={\int }_a^{b}f(x)dx$

This approach leads to the exact value of the area under f(x).

3.0Area Under the Normal Curve

In probability and statistics, the area under the normal curve is of great significance. The normal distribution curve, also known as the bell curve, is symmetric about the mean. The total area under the curve is equal to 1. The area under the standard normal curve (where the mean $\mu =0$ and standard deviation $\sigma =1$ is used to calculate probabilities and z-scores in statistical problems.

For any interval [a, b] under the standard normal curve, the area can be computed using the standard normal distribution table or software tools.

4.0Area Under Two Curves

The concept of finding the area under two curves involves calculating the area of the region enclosed between two curves f(x) and g(x). If $f(x)\ge g(x)$ over the interval [a, b], the area between the two curves is given by:

$\text{ Area }={\int }_a^b[f(x)-g(x)]dx$

This formula subtracts the area under g(x) from the area under f(x).

5.0Practical Applications

Calculating the area under a curve has numerous real-world applications, such as finding the displacement of an object (using velocity-time graphs), computing probabilities in statistics, and determining the work done by a force in physics. Understanding the fundamental principles and mastering problem-solving techniques is essential for tackling competitive exams like JEE Mains.

6.0Solved Examples on Area Under Curve

Example 1: Find the area of the region bounded by the curve $y=x^2$ and the line y = 4.

Solution: 

The intersection points of the curve $y=x^2$ and the line y = 4 occur when $x^2=4$ , i.e., x = –2 and x = 2.

The area under $y=x^2$ from x = –2 to x = 2 is: 

$\text{ Area }={\int }_a^b[f(x)-g(x)]dx$

Here $\mathrm{f}(\mathrm{x})=4;\mathrm{g}(\mathrm{x})=x^2$

$\text{ Area }={\int }_{-2}^{2}4-x^{2}dx$

Calculating this integral:

$\text{ Area }={\left[4x-\frac{x^3}{3}\right]}_{-2}^2$

$\text{ Area }=\left(8-\frac{8}{3}\right)-\left(-8+\frac{8}{3}\right)$

$\text{ Area }=\frac{32}{3}\text{ square units }$

Hence, area of the region bounded by the curve $y=x^2$ and the line y = 4 is

$\frac{32}{3}\text{ square units }.$

Example 2: Find area enclosed between $y=\sin x\&x$ -axis as x varies from 0 to $\frac{3\pi }{2}$ . 

 Solution:

$\text{ Area }={\int }_0^{\pi }\sin xdx+\left|{\int }_{\pi }^{3\pi /2}\sin xdx\right|$

$=[-\cos x{]}_0^{\pi }+\left|[-\cos x{]}_{\pi }^{\frac{3\pi }{2}}\right|$

=(1+1)+|(0-1)|

= 2 + 1 = 3 square units.

Hence, area enclosed between $y=\sin x\&$ x -axis is 3 square units.

Example 3: The area of the region bounded by the curves $y=|x-2|$, $x=1$, x=3 and the x-axis is

(A) 4 (B) 2 (C) 3 (D) 1

Ans. (D)

Solution:

Required area $={\int }_1^3|x-2|dx$

    $={\int }_1^2(2-x)dx+{\int }_2^3(x-2)dx$

        $={\left[2x-\frac{x^2}{2}\right]}_1^2+{\left[\frac{x^2}{2}-2x\right]}_2^3$

        $={\left[2x-\frac{x^2}{2}\right]}_1^2+{\left[\frac{x^2}{2}-2x\right]}_2^3$

        $=\frac{1}{2}+\frac{1}{2}=1$  square units.

Hence, area of the region bounded by the curves y=|x-2|, $x=1,x=3$ and the x-axis is 1 square units.

Example 4: Find the area in the first quadrant bounded by y=4 x^2, x=0, y=1 and y=4 .

Solution:

Required area

$={\int }_1^{4}xdy={\int }_1^4\frac{\sqrt{y}}{2}=dy$ $=\frac{1}{3}\left[4^{3/2}-1\right]$

      $=\frac{1}{3}[8-1]$

$=\frac{7}{3}=2^{\frac{1}{3}}$ square units.

Hence, area in the first quadrant bounded by $y=4x^2,x=0,y=1$ and $y=4\text{ is }2\frac{1}{3}$  square units.

Example 5: The area of figure bounded by $y=e^x,y=e^{-x}$ and the straight line x = 1 is

(A) $e+\frac{1}{e}$ (B) $e-\frac{1}{e}$ (C) $e+\frac{1}{e}-2$ (D) $e+\frac{1}{e}+2$

Ans. (C)

Solution:

Given equations of curves $y=e^x;y=e^{-x}$ and straight line x = 1 

We know that area of the figure bounded by the curves and straight line  

$={\int }_0^1\left(e^x-e^{-x}\right)dx$

$={\left[e^x+e^{-x}\right]}_0^1$

$=e+\frac{1}{e}-2$

Hence, area of figure bounded by $y=e^x,y=e^{-x}$ and the straight line x=1 is $e+\frac{1}{e}-2$.

Example 6: Area enclosed between the curves $y=ex\cdot \ln x$ and $y=\frac{\ln x}{ex}$

Solution:

Solving $e{x}^{\ln x=\frac{\ln x}{ex}}$

$\ln x\left(e^2{x}^2-1\right)=0$

$x=1\text{ or }x=\frac{1}{e}$

Where $x=1;y_1=0;y_2=0$

$x=\frac{1}{e};y_1=-1;y_2-1$

Also examine the increasing and decreasing behaviour of the curve

$A={\int }_{1/e}^1\left(\frac{\ln x}{ex}-ex\ln x\right)dx=\frac{e}{4}-\frac{5}{4e}$

Hence, area enclosed between the curves $y=ex\cdot \ln xandy=\frac{e}{4}-\frac{5}{4e}$ is.

Example 7: Area of the loop a y^2=x^2(a-x), a>0

Solution:

$a{y}^2=x^2(a-x),a\ge 0$

Cuts x-axis at 

$0=x^2(a-x)$

$x^2(a-x)=0$

$\therefore x=0;x=a$

cuts y-axis at 

$a{y}^2=0$

y = 0

$A={\int }_0^a\left(\frac{x\sqrt{a-x}}{\sqrt{a}}-\left(\frac{-x\sqrt{a-x}}{\sqrt{a}}\right)\right)dx$

OR $A=2{\int }_0^a\frac{x}{\sqrt{a}}(\sqrt{a-x})dx=\frac{8}{15}{a}^2$

Hence, Area of the loop $a{y}^2=x^2(a-x)$, $a>0\text{ is }\frac{8}{15}{a}^2$ square units.

Example 8: Find the area bounded by the regions $y^2\ge x,x>-$ & curve $x^2+y^2\le 2$

Solution:

Common region is given by the diagram  

If area of region O A B= $\lambda$

then area of O C D= $\lambda$

Because $y=\&x=-\sqrt{y}$

will bound same area with x & y axes respectively.

$y=\sqrt{x}\Rightarrow y^2=x$

$x=\Rightarrow x^2=y$ and hence both the curves are  symmetric with respect to the line y = x Area of first quadrant

$OBC=\frac{\pi r^2}{4}=\frac{\pi }{2}(\because r=\sqrt{2})$

Area of region $OCA=-\lambda$

Area of shaded region = $\left(\frac{\pi }{2}-\lambda \right)+\lambda =\frac{\pi }{2}$ square units.

Hence, area bounded by the regions $y^2\ge x,x>-$ & curve $x^2+y^2\le 2$ is $\frac{\pi }{2}$ square units.

This type of question is frequently asked in JEE Mains and Advanced exams, making it essential for students to practice and understand the concept thoroughly.

7.0Practice Problems on Area Under Curve

1. Find the area of the region enclosed by the parabola y = x² and the line y = x + 4.
2. Determine the area between y = sinx and y = cos x over the interval $[0,\pi ]$ .
3. Solve for the area between y = e^x and y = 2e^x over the interval [0, 1].

8.0Sample Questions on Area Under Curve

1. What is the area under a curve?

Ans: The area under a curve refers to the region bounded by the curve, the x-axis, and the specified limits on the x-axis. It is typically measured using definite integrals in calculus. If a function f(x) is defined and continuous over an interval [a, b], the area under the curve f(x) from x = a to x = b can be expressed as:

$\text{ Area }={\int }_a^{b}f(x)dx$

2. How do you calculate the area under a curve using calculus?

Ans: To find the area under a curve using calculus, you use definite integrals. The integral ${\int }_a^{b}f(x)dx$ represents the area under f(x) over the interval [a, b]. If f(x) is above the x-axis, the area is positive; if f(x) is below the x-axis, the area is negative.

3. What is the formula for finding the area between two curves?

Ans: The area between two curves f(x) and g(x) over the interval [a, b] is calculated as:

$\text{ Area }={\int }_a^b[f(x)-g(x)]dx$

This formula subtracts the area under the lower curve g(x) from the area under the upper curve f(x).

4. What is the relationship between definite integrals and area under curves? 

Ans: Definite integrals are used to compute the area under curves. The definite integral ${\int }_a^{b}f(x)dx$ gives the net area between f(x) and the x-axis from x = a to x = b. If f(x) is positive, the area is above the x-axis; if f(x) is negative, the area is below the x-axis.