Question

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

Answer 1

Plotting $y=3x+2$ in the graph,

$x$	$0$	$\frac{-2}{3}=-0.66$
$y$	$2$	$0$

Area Required$=$Area $△ACB+$Area of $△ADE$

Area of $△ACB={\int }_{-1}^{\frac{-2}{3}}(3x+2)dx$

Since area $ACB$ is below $x-$axis, it will come negative,

Hence, we take modulus

Area $ACB=\left|{\int }_{-1}^{\frac{-2}{3}}(3x+2)dx\right|$

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

$=|[\frac{3}{2}\times \frac{4}{9}-\frac{4}{3}]-[\frac{3}{2}-2]|$

$=|\frac{-2}{3}+\frac{1}{2}|=\left|\frac{-1}{6}\right|=\left|\frac{1}{6}\right|$

Area $ADE=\left|{\int }_{\frac{-2}{3}}^{-1}(3x+2)dx\right|$

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

$=[\frac{3}{2}+2]-[\frac{2}{3}-\frac{4}{3}]$

$=\frac{7}{2}+\frac{2}{3}=\frac{25}{6}$

Thus, Required area$=$Area$ACB+$Area$ADE$

$=\frac{1}{6}+\frac{25}{6}=\frac{26}{6}=\frac{13}{3}$sq.units