Question

Obtain the area enclosed by the curve $x+2y-8=0$, $x$-axis & lines $x=1$ & $x=5$ by using integration.

Answer 1

Given

$x+2y-8=0$

$2y=8-x$

$y=\frac{8-x}{2}$

and x=1 & x=5

The required are is $A={\int }_1^5\frac{8-x}{2}dx$

$=\frac{1}{2}{\int }_1^{5}8-xdx$

$=\frac{1}{2}[8x-\frac{x^2}{2}{]}_1^5$

$=\frac{1}{2}[8\times 5-8\times 1-(\frac{5^2}{2}-\frac{1}{2}\left)\right]$

$=\frac{1}{2}[40-8-(\frac{25}{2}-\frac{1}{2}\left)\right]$

$=\frac{1}{2}[32-\frac{24}{2}]$

$=\frac{1}{2}(32-12)$

$=\frac{1}{2}\left(20\right)=10$ sq. units