Question

Find the area bounded by lines $y=4x+5,y=5-x\&4y=x+5$

Answer 1

We have $y=4x+5$ ..(1)
$y=5-x$ ..(2)
$4y=x+5$ . .(3)
Solving(1) and (2)
Subtract (2) from(1)
$x=0\Rightarrow y=5$
Hence ,$A(0,5)$ , is the point of intersection of lines (1) and (2)
Solving (1) and (3)
Multiply (1) by 4 and subtract from (3)
$x-16x+5-20=0\Rightarrow x=-1\Rightarrow y=1$
Hence$B(-1,1)$ is the point of intersection of lines (1) and (2)
Solving(2) and (3)
$y=2\Rightarrow x=3$
Hence $C\left(3.2\right)$ is the point of intersection of line (2) & (3)
So, the area bounded by curves in the diagram below:

Required area is the sum of area bounded by line $y=4x+5\text{and}4y=x+5$ when $x$ varies from $-1\text{to}0$ and the area bounded by line $y=5-x\text{and}4y=x+5$ when $x$ varies from $0\text{to}3$, therefore
$Area={\int }_{-1}^0(y_{up}-y_{down})dx+{\int }_0^3(y_{up}-y_{down})dx$
$={\int }_{-1}^0(4x+5-\frac{x+5}{4})dx+{\int }_0^3(5-x-\frac{x+5}{4})dx$
[$\because x$ varies from-1 to 2]
$={[\frac{4x^2}{2}+5x-\frac{x^2}{8}-\frac{5x}{4}]}_{-1}^0+{[5x-\frac{x^2}{2}-\frac{x^2}{8}-\frac{5x}{4}]}_0^3$
$[\because {\int }_a^b{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_a^b]$
$=[0-(2-5-\frac{1}{8}+\frac{5}{4})]+[15-\frac{9}{2}-\frac{9}{8}-\frac{15}{4}]=[\frac{-16+40+1-10}{8}+\left[\frac{120-36-9-30}{8}\right]]$
$=\left[\frac{15}{8}\right]+\left[\frac{45}{8}\right]$
$=\frac{60}{8}$
$=\frac{15}{2}\text{sq units}$
Hence , the required area is $\frac{15}{2}\text{sq. units}$