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Area Between Two Curves in Calculus

Area Between Two Curves in Calculus

Edited By Komal Miglani | Updated on Oct 15, 2024 01:47 PM IST

Area bounded by curves and axis is one of the important parts of Calculus, which applies to measuring the change in the function at a certain point. Mathematically, it forms a powerful tool by which slopes of functions are determined, the maximum and minimum of functions found, and problems on motion, growth, and decay, to name a few. These integration concepts have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

Area Between Two Curves in Calculus
Area Between Two Curves in Calculus

Area Bounded by Curve and Axes

If the function $f(x) ≥ 0 ∀ x ∈ [a, b]$ then $\int_a^{\infty} f(x) d x$ represents the area bounded by $y = f(x), x-$axis and lines $x = a$ and $x = b$.

If the function $f(x) ≤ 0 ∀ x ∈ [a, b]$, then the area by bounded 4y = f(x), x-4axis and lines 4x = a$ and $x = b$ is $\int_a^b f(x) d x$.

Area along Y-axis

The area by bounded $x = g(y)$ [with $g(y)>0$], $y$-axis and the lines $y = a$ and $y = b$ is $\int_a^b x d y=\int_a^b g(y) d y$

Area of Piecewise Function

If the graph of the function $f(x)$ is of the following form, then

then $\int_a^b f(x) d x$ will equal $A_1-A_2+A_3-A_4$ and not $A_1+A_2+A_3+A_4$.

If we need to evaluate $A_1+A_2+A_3+A_1$ (the magnitude of the bounded area) we will have to calculate

$ \underbrace{\int_a^x f(x) d x}_{\mathrm{A}_1}+\underbrace{\left|\int_x^y f(x) d x\right|}_{\mathrm{A}_2}+\underbrace{\int_5^z f(x) d x}_{\mathrm{A}_3}+\underbrace{\left|\int_z^b f(x) d x\right|}_{\mathrm{A}_4} $

The area bounded by curve when curve intersects $X$-axis

The graph $y=f(x) \forall x \in[a, b]$ intersects $x-a x i s$ at $x=c$.

If the function $f(x) \geq 0 \forall x \in[a, c]$ and $f(x) \leq 0 \forall x \in[c, b]$ then area bounded by curve and $x$-axis, between lines $x=a$ and $x=b$ is

$ \int_a^b|f(x)| d x=\int_a^c f(x) d x-\int_c^b f(x) d x$


Area Bounded by Two Curves:

Area bounded by the curves $y=f(x), y=g(x) $ and the lines $ x = a$ and $x = b$, and it is given that $f(x) ≤ g(x). $


From the figure, it is clear that,

Area of the shaded region = Area of the region $ABEF$ - Area of the region $ABCD$

$\int_a^b g(x) d x-\int_a^b f(x) d x=\int_a^b(\underbrace{g(x)}_{\begin{array}{c}\text { upper } \\ \text { curve }\end{array}}-\underbrace{f(x)}_{\begin{array}{c}\text { lower } \\ \text { curve }\end{array}}) d x$

Area Bounded by Curves When Intersects at More Than One Point:

Area bounded by the curves $y = f(x), y = g(x)$ which intersect each other in the interval $[a, b]$

First find the point of intersection of these curves $y = f(x)$ and $y = g(x)$ by solving the equation $f(x) = g(x)$, let the point of intersection be $x = c $

Area of the shaded region

$=\int_a^c\{f(x)-g(x)\} d x+\int_c^b\{g(x)-f(x)\} d x$

When two curves intersects more than one point

Area bounded by the curves $y=f(x), y=g(x)$ which intersect each other at three points at $x = a, x = b$ and $x = c. $

To find the point of intersection, solve $f(x) = g(x). $

For $x ∈ (a, c), f(x) > g(x)$ and for $x ∈ (c, b),g(x) > f(x).$

Area bounded by curves,

$\begin{aligned} A & =\int_a^b|f(x)-g(x)| d x \\ & =\int_a^c(f(x)-g(x)) d x+\int^b(g(x)-f(x)) d x\end{aligned}$

Recommended Video Based on Area Between two Curves

Solved Examples

Example 1: The area of the region (in sq. units) bounded by the curves $y=|x-1|$ and $y=3-|x|$ is
1) $4$
2) $3$
3) $6$
4) $2$

Solution:

The area between the curve $y=f(x), x$ axis and two ordinates at the point $x=a$ and $x=b(b>a)$ is given by $A=\int_a^b f(x) d x=\int_a^b y d x$

$ y=(x-1)$ and $y=3-(x)$


Example 2:The area of the region enclosed by the curves $y=x, x=e, y=1 / x$ and the positive $x$-axis is
1) $3 / 2$ square units
2) $5 / 2$ square units
3) $1 / 2$ square units
4) $1$ square units

Solution:

$\begin{aligned} & \int_0^1 x d x+\int_1^e \frac{1}{x} d x \\ & \Rightarrow\left[\frac{x^2}{2}\right]_0^1+[\log x]_1^e \\ & =\frac{1}{2}+\ln e \\ & =\frac{1}{2}+1=\frac{3}{2} \text { sq units }\end{aligned}$

Example 3:

The area bounded by the curves $y=\cos x$ and $y=\sin x$ between the ordinates $x=0$ and $x=\frac{3 \pi}{2}$ is
1) $
4 \sqrt{2}-2
$

2) $
4 \sqrt{2}+2
$

3) $
4 \sqrt{2}-1
$

4) $
4 \sqrt{2}+1
$

Solution

Area along $x$ axis -
Let $y_1=f_1(x)$ and $y_2=f_2(x)$ be two curve then area bounded between the curves and the lines $x=a$ and $x=b$ is

$
\left|\int_a^b \Delta y d x\right|=\left|\int_a^b\left(y_2-y_1\right) d x\right|
$


$\text { Where } \Delta y=f_2(x)-f_1(x)$

$\begin{aligned} & \text { Area }=\int_0^{\frac{\pi}{4}}(\cos x-\sin x) d x+\int_{\frac{\pi}{4}}^{\frac{5 \pi}{4}}(\sin x-\cos x) d x+\int_{\frac{5 \pi}{4}}^{\frac{3 \pi}{2}}(\cos x-\sin x) d x \\ & =[\sin x+\cos x]_0^{\frac{\pi}{4}}+[-\cos -\sin x]_{\frac{5 \pi}{4}}^{\frac{5 \pi}{4}}+[\sin x+\cos x]_{\frac{5 \pi}{4}}^{\frac{3 \pi}{2}} \\ & =\sqrt{2}^{-1}+[\sqrt{2}+\sqrt{2}]+1(-1)+\sqrt{2} \\ & =4 \sqrt{2}-2\end{aligned}$

Example 4: The area (in sq. units) of the region described by $\mathrm{A}=\left\{(x, y) \mid y \geq x^2-5 x+4, x+y \geq 1, y \leq 0\right\}$ is :
1) $\frac{7}{2}$
2) $\frac{19}{6}$
3) $\frac{13}{6}$
4) $\frac{17}{6}$

Solution

Area along $x$ axis -
Let $y_1=f_1(x)$ and $y_2=f_2(x)$ be two curve then area bounded between the curves and the lines $x=a$ and $x=b$ is

$
\left|\int_a^b \Delta y d x\right|=\left|\int_a^b\left(y_2-y_1\right) d x\right|
$


\text { Where } \Delta y=f_2(x)-f_1(x)

Point of intersection of $y=x^2-5 x+4$ and $x+y=1$ are $x=1,3$ at $x=1, y=0$;

$
x=3, y=-2
$

$\begin{aligned} & \text { Req Area }=\text { Area } \triangle A B C+\left|\int_3^4\left(x^2-5 x+4\right) d x\right| \\ & =\frac{1}{2} \times 2 \times 2+\left[\frac{x^3}{3}-\frac{5 x^2}{2}+4 x\right]_3^4 \\ & =\frac{19}{6}\end{aligned}$

Example 5: Area bounded (in sq. units) by the curves $4 y=\left|x^2-4\right|$ and $y+|x|=7$, is equal to
1) $32$
2) $16$
3) $4$
4) $8$

Solution

\begin{aligned}
&\text { If we have two functions intersection each other.First find the point of intersection. Then integrate to find area }\\
&\int_o^a[f(x)-9(x)] d x
\end{aligned}

$\begin{aligned} \text { Required area } & =2\left(\int_0^2\left(7-x-\left(\frac{4-x^2}{4}\right)\right) d x+\int_2^4\left(7-x-\left(\frac{x^2-4}{4}\right)\right) d x\right) \\ = & 2\left(\int_0^2\left(6-x-\frac{x^2}{4}\right) d x+\int_2^4\left(8-x-\frac{x^2}{4}\right) d x\right) \\ = & 2\left(6 x-\frac{x^2}{2}+\frac{x^3}{12}\right)_0^2+2\left(8 x-\frac{x^2}{2}-\frac{x^3}{12}\right)_2^4=32 \text { sq.units }\end{aligned}$

Summary

Area bounded by two curves involves integrating the given function over a specified interval. When the function is non-negative over the interval, the integral directly provides the area. Understanding these principles allows us to handle a variety of scenarios, including the area between two curves.

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