Area Under the Curve

The area under the curve is a concept in integral calculus that quantifies the total area enclosed by a curve, the x-axis, and specified vertical boundaries on the graph of a function. This area is calculated using definite integrals and represents the accumulation of a quantity described by the function over a given interval.

1.0Mathematical Definition

For a continuous function f(x) defined on the interval [a, b], the area under the curve 

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

from x = a to x = b is given by the definite integral: 

2.0Area Under the Curve Definition

The area under the curve is a fundamental concept in integral calculus that quantifies the total area between the graph of a function and the x-axis over a specified interval. Mathematically, it is determined using definite integrals.

Mathematical Representation:

If f(x) is a continuous function defined on the interval [a, b], the area under the curve 

Area= ${\int }_a^{b}f\left(x\right)dx$ from x = a to x = b is given by the definite integral: .

3.0Area Under the Curve – Between a Curve and Coordinate Axis

1. Area under the Curve and x- axis (Vertical Strips)

Area of the strip = y.dx

Area bounded by the curve, the x-axis and the ordinate at $x=a$ and $x=b$ is given by 

$A={\int }_a^{b}ydx$, where  $y=f\left(x\right)$ lies above the x-axis and _b>a

Here vertical strip of thickness $\mathrm{d}x$ is considered at distance $x$_.

If $y=f\left(x\right)$  lies completely below the $x-$axis, then $A=\left|{\int }_a^{b}ydx\right|$

If curve crosses the _x -axis at _x=c , then _{A=\left|\int_{aydx}^{c}\right|+\int_{c}^{b}ydx}

4.0Area under the Curve and y- axis (Horizontal Strips)

Area of the strip = x.dy

Graph of x = g(y)

If $g\left(y\right)\ge 0$ for $y\in \left[c,d\right]$  then area bounded by curve $x=g\left(y\right)$ and y-axis between abscissa $y=c$ and is ${\int }_{y=c}^{d}g\left(y\right)\mathrm{d}y$

If $g\left(y\right)\le 0$ for $y\in \left[c,d\right]$ then area bounded by curve $x=g\left(y\right)$ and y-axis between abscissa  $y=c$ and $y=d$ is ${\int }_{y=c}^{d}g\left(y\right)\mathrm{d}y$

Note:

General formula for area bounded by curve  $x=g\left(y\right)$  and y-axis between abscissa  $y=c$  and  $y=d$ is  ${\int }_{y=c}^d\left|g\left(y\right)\right|\mathrm{d}y$ .

5.0Area under the curve - Symmetric Area

If the curve is symmetric in all four quadrants, then 

Total area = 4 (Area in any one of the quadrants).

6.0Area Under a curve - Between Two Curves       

1. Area bounded by two curves $y=f\left(x\right)\&y=g\left(x\right)$

such that  $f\left(x\right)>g\left(x\right)$ is $A={\int }_{x_1}^{x_2}\left[f\left(x\right)-g\left(x\right)\right]\mathrm{d}x$

where $x_1$ and $x_2$ are roots of equation  $f\left(x\right)=g\left(x\right)$

2. In case horizontal strip is taken we have

$A={\int }_{y_1}^{y_2}\left[f\left(y\right)-g\left(y\right)\right]dy$

Where  $y_1\&y_2$  are roots of equation $f\left(y\right)=g\left(y\right)$

3. If the curves $y=f\left(x\right)$ and  $y=g\left(x\right)$ intersect at $x=c$, then required area

$A={\int }_a^c\left(g\left(x\right)-f\left(x\right)\right)dx+{\int }_c^b\left(f\left(x\right)-g\left(x\right)\right)dx={\int }_a^b\left|f\left(x\right)-g\left(x\right)\right|dx$ 

Note: Required area must have all the boundaries indicated in the problem.

7.0Standard Areas (To be Remembered)

1. Area bounded by parabolas  $y^2=4ax;x^2=4by,a>0;b>0$ is $\frac{16ab}{3}$

Intersection point:

$\left\{y^2=4ax\right\}$

$\left\{x^2=4by\right\}$

$x^2=4b\sqrt{4ax}$

$x^4=64b^{2}ax$

$x=0orx=4b^{\frac{2}{3}}{a}^{\frac{1}{3}}$

$A={\int }_0^{4b^{\frac{2}{3}}{a}^{\frac{1}{3}}}\left(\sqrt{4ax}-\frac{x^2}{4b}\right)dx=\frac{16ab}{3}$

2. Whole area of the ellipse,  $x^2/a^2+y^2/b^2=1$ is  $\pi ab$ sq. units.
3. Area included between the parabola $y^2=4ax$ & the line $y=mx$ is  $8a^2/3m^3$ sq. units.
4. The area of the region bounded by one arch of sin $ax$ (or cos $ax$) and $x$-axis is $2/a$ sq. units. 
5. Average value of a function  $y=f\left(x\right)$ over an interval $a\le x\le b$ is defined as:

$y\left(av\right)=\frac{1}{b-a}{\int }_a^{b}f\left(x\right)\mathrm{d}x$

6. If $y=f\left(x\right)$ is a monotonic function in $\left(a,b\right)$, then the area bounded by the ordinates at $x=a$,

$x=b,y=f\left(x\right)andy=f\left(c\right)[wherec\in \left(a,b\right)]isminimumwhenc=\frac{a+b}{2}$.

8.0Solved Examples on Area Under the Curve

Example 1: Find area bounded by $x=1,x=2,y=x^2,y=0$. 

Solution:

$A={\int }_1^2\left(x^2-0\right)dx$

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

$=\frac{8-1}{3}=\frac{7}{3}$

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

Solution:

Required area = ${\int }_1^{4}xdy={\int }_1^4\frac{\sqrt{y}}{2}\mathrm{d}y$

$=\frac{1}{2}\frac{.2}{3}.{\left(y^{\frac{3}{2}}\right)}_1^4$

$=\frac{1}{3}\left[4^{\frac{3}{2}}-1\right]$

$=\frac{1}{3}\left[8-1\right]=\frac{7}{3}=2\frac{1}{3}$ sq. units.

Example 3: Find the area enclosed between $y=\sin x;y=\cos x$ and y-axis in the 1^st quadrant

Solution:

$A={\int }_0^{\frac{\pi }{4}}\left(\cos x-\sin x\right)dx$

$={\left[\sin x+\cos x\right]}_0^{\frac{\pi }{4}}=\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-\left(0+1\right)$

$=\sqrt{2}-1$

Example 4: Find the area bounded by the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$

Solution:

Area bounded by ellipse in first quadrant $={\int }_0^3\frac{2}{3}\sqrt{9-x^2}dx=\frac{3\pi }{2}$

∵ Curve is symmetrical about all four quadrants

∴ Total area = 4 (Area in any one of the quadrants)

$=4\left(\frac{3\pi }{2}\right)=6\pi$

Example 5: Compute the area of the figure bounded by the parabolas _​​

Solution:

Solving the equations $x=-2y^2,x=1-3y^2$, 

we find that ordinates of the points of intersection 

of the two curves as $y_1=-1,y_2=1$

The points are (–2, –1) and (–2, 1).

The required area

$2{\int }_0^1\left(x_1-x_2\right)\mathrm{d}y=2{\int }_0^1\left[\left(1-3y^2\right)-\left(-2y^2\right)\right]\mathrm{d}y=2{\int }_0^1\left(1-y^2\right)dy$

$=2{\left[y-\frac{y^3}{3}\right]}_0^1=\frac{4}{3}$ sq.units.

9.0Practice Problems on Area Under the Curve

1. Find the area bounded by y = x² + 2 above  x-axis between x = 2 and x = 3.

2. Find the area bounded by the curve y = cos x  and the x-axis  from  x = 0 to x = 2π..

3. Find the area bounded by x = 2, x = 5, y = x², y = 0.

4. Find the area bounded by  y = x² and y = x.

5. A figure is bounded by the curves $y=\left|\sqrt{2}\sin \frac{\pi x}{4}\right|$ , y = 0, x = 2 and x = 4. At what angles to the positive x-axis straight lines must be drawn through (4, 0) so that these lines partition the figure into three parts of the same area.