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 enclosed by the lines

y = 3 x + 2

,

x

-axis and ordinates

x = - 1

and

x = 1

.

Accepted Answer

Given, $y = 3 x + 2,$ $x$ -axis, $x = - 1$ and $x = 1$ . Draw the lines on coordinate axes.

Finding intersection point of $y = 3 x + 2$ and $x$ -axis,

At $x$ -axis, $y = 0$

So, $3 x + 2 = 0$

$\Rightarrow x = - \frac{2}{3}$

Required area $= |\int_{- 1}^{- \frac{2}{3}} y d x| + \int_{- \frac{2}{3}}^{1} y d x$

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

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

$= |\frac{3 {(- \frac{2}{3})}^{2}}{2} + 2 (- \frac{2}{3}) - (\frac{3}{2} - 2)| + (\frac{3}{2} + 2 - \frac{3 {(- \frac{2}{3})}^{2}}{2} - 2 (- \frac{2}{3}))$

$= |\frac{2}{3} - \frac{4}{3} - \frac{3}{2} + 2| + (\frac{3}{2} + 2 - \frac{2}{3} + \frac{4}{3})$

$= |\frac{- 2}{3} + \frac{1}{2}| + (\frac{7}{2} + \frac{2}{3})$

$= |\frac{- 4 + 3}{6}| + (\frac{21 + 4}{6})$

$= |\frac{- 1}{6}| + (\frac{25}{6})$

$= \frac{1}{6} + \frac{25}{6}$

$= \frac{26}{6}$

$= \frac{13}{3}$ square units.

Find the area enclosed by the lines $y = 3 x + 2$ , $x$ -axis and ordinates $x = - 1$ and $x = 1$ .

Important Questions on Application of Integrals

Learn and Prepare for any exam you want !

Find the area enclosed by the lines y=3x+2, x-axis and ordinates x=-1 and x=1.

Important Questions on Application of Integrals

Learn and Prepare for any exam you want !

Find the area enclosed by the lines $y = 3 x + 2$ , $x$ -axis and ordinates $x = - 1$ and $x = 1$ .