Name: Area Enclosed by a Curve and a Line: Illustration
Uploaded: 2019-07-19
Duration: 3 min 35 s
Description: How would you find the area enclosed between a parabola and a straight line? Watch the video to see the method as an application of definite integrals.

Question

Find the area of the region bounded by the curve $y = x^{2}$ , the $x$ -axis, and the lines $x = 1$ and $x = 3$ .

Accepted Answer

Given the boundaries of the area to be found are.

The curve $y = x^{2}$

The $x$ -axis

$x = 1$ (a line parallel to $y$ -axis)

$x = 3$ (a line parallel to $y$ -axis)

As per the given boundaries,

The curve $y = x^{2}$ , has only the positive numbers as $x$ has even power, so it is about the $y$ -axis equally distributed on both sides.

$x = 1$ and $x = 3$ are parallel to $y$ -axis at of $1$ and $3$ units respectively from the $y$ -axis.

The four boundaries of the region to be found are,

Point $A$ , where the curve $y = x^{2}$ and $x = 3$ meet

Point $B$ , where the curve $y = x^{2}$ and $x = 1$ meet

Point $C$ , where the $x$ -axis and $x = 1$ meet i.e. $C (1, 0)$ .

Point $D$ , where the $x$ -axis and $x = 3$ meet i.e. $D (3, 0)$

Area of the required region $=$ Area of $A B C D$

Area of $ABCD = \int_{1}^{3} x^{2} d x$

$= {[\frac{x^{3}}{3}]}_{1}^{3}$ (on integration)

$= {[\frac{x^{3}}{3}]}_{1}^{3}$

$= (\frac{3^{3}}{3} - \frac{1^{3}}{3})$

$= (\frac{27}{3} - \frac{1}{3})$

$= \frac{26}{3}$

The Area of the required region $A B C D$ is $\frac{26}{3}$ sq.units.

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