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 line

y = 3 x + 2, X - a x i s

and the ordinates

x = - 1

and

x = 1

.

Accepted Answer

Given that,

We have to find the area of the region bounded by the line $y = 3 x + 2, X - a x i s$ and the ordinates $x = - 1$ and $x = 1$ .

We have, $y = 3 x + 2$ .

At $x = 0, y = 2$ .

At $x = 1, y = 5$ .

At $x = \frac{- 2}{3}, y = 0$ .

At $x = - 1, y = - 1$ .

Let us plot the above points to get the graph of $y = 3 x + 2$ . Also plot $x = - 1, x = 1$ graph.

The area of the region is given by:

$A =$ Area of $A C B +$ Area of $A D E$ .

$= \int_{- 1}^{\frac{- 2}{3}} (3 x + 2) d x + \int_{\frac{- 2}{3}}^{1} (3 x + 2) d x = |{[\frac{3 x^{2}}{2} + 2 x]}_{- 1}^{\frac{- 2}{3}}| + {[\frac{3 x^{2}}{2} + 2 x]}_{\frac{- 2}{3}}^{1}$

Since $A C B$ is below $x - a x i s$ we will take modulus.

$= |[\frac{3 {(\frac{- 2}{3})}^{2}}{2} + 2 \times (\frac{- 2}{3}) - (\frac{3 \times {(- 1)}^{2}}{2} + 2 \times - 1)]| + [\frac{3 \times 1^{2}}{2} + 2 \times 1 - (\frac{3 {(\frac{- 2}{3})}^{2}}{2} + 2 \times (\frac{- 2}{3}))] = |\frac{2}{3} - \frac{4}{3} - \frac{3}{2} + 2| + \frac{3}{2} + 2 - \frac{2}{3} + \frac{4}{3} = |\frac{- 1}{6}| + \frac{25}{6} \Rightarrow \frac{1}{6} + \frac{25}{6} = \frac{13}{3} sq . units$

Therefore, the area of the region bounded by $y = 3 x + 2, X - a x i s, x = - 1, x = 1$ is $\frac{13}{3} sq . units$ .

Find the area of the region bounded by the line $y = 3 x + 2, X - a x i s$ and the ordinates $x = - 1$ and $x = 1$ .

Important Questions on Applications of Integrals

Learn and Prepare for any exam you want !

Find the area of the region bounded by the line y=3x+2, X-axis and the ordinates x=-1 and x=1.

Important Questions on Applications of Integrals

Learn and Prepare for any exam you want !

Find the area of the region bounded by the line $y = 3 x + 2, X - a x i s$ and the ordinates $x = - 1$ and $x = 1$ .