Question

Using integration find the area of triangle whose sides are given by the equation $y=x+1,y=3x+1,x=5$

Answer 1

$y=x+1\left(\text{Given}\right)$

$x$	$0$	$1$
$y$	$1$	$2$

$y=3x+1\left(\text{Given}\right)$

$x$	$0$	$1$
$y$	$1$	$4$

$x=5\left(\text{Given}\right)$

Thus,

Required area $=$ Area of $△ABC$

Now,

From fig.,

Area of $△ABC=ar\left(OABD\right)-ar\left(OACD\right)={\int }_0^5(3x+1)dx-{\int }_0^5(x+1)dx={\int }_0^5(3x+1-x-1)dx={\int }_0^{4}2xdx={\left[\frac{2x^2}{2}\right]}_0^5=({\left(5\right)}^2-{\left(0\right)}^2)=25\text{sq. units}$

Hence the area of $△ABC$ is $25\text{sq. units}$.