#### Using integration, find the area of the region bounded by the line 2y = 5x + 7, x-axis and the lines x = 2 and x = 8.

The equation of the line is 2y = 5x + 7

Now, we need point of the line, so we substitute x = 0 and then y = 0

$\\Put \: \: x=0 \Rightarrow 2 y=5(0)+7\\ \Rightarrow \mathrm{y}=\frac{7}{2}\\ Put\: \: y=0\\ \\\Rightarrow 2(0)=5 x+7 \\\Rightarrow 5 x=-7 \\\Rightarrow x=-\frac{7}{5}$

$\\Hence \left(0, \frac{7}{2}\right) and \left(-\frac{7}{5}, 0\right) are the required two points to draw the line 2 \mathrm{y}=5 \mathrm{x}+7$

The other two x = 2 and x = 8 are straight lines parallel to Y axis.

On plotting all the above lines.

$\\\text{We have to find area under} 2 \mathrm{y}=5 \mathrm{x}+7\\ that is \mathrm{y}=1 / 2(5 \mathrm{x}+7) from 2 to 8\\ \Rightarrow y=1 / 2(5 x+7)\\ Integrate from 2 to 8\\ \Rightarrow \int_{2}^{8} \mathrm{ydx}=\frac{1}{2} \int_{2}^{8}(5 \mathrm{x}+7) \mathrm{dx} \\\Rightarrow \int_{2}^{8} \mathrm{ydx}=\frac{1}{2}\left[\frac{5 \mathrm{x}^{2}}{2}+7 \mathrm{x}\right]_{2}^{8} \\\Rightarrow \int_{2}^{8} \mathrm{ydx}=\frac{1}{2}\left[\left(\frac{5(8)^{2}}{2}+7(8)\right)-\left(\frac{5(2)^{2}}{2}+7(2)\right)\right] \\\Rightarrow \int_{2}^{8} \mathrm{ydx}=\frac{1}{2}[(160+56)-(10+14)]$

\\\begin{aligned} &\Rightarrow \int_{2}^{8} \mathrm{ydx}=\frac{1}{2}(192)\\ &\Rightarrow \int_{2}^{8} \mathrm{ydx}=96\\ &\text { Hence the area bounded by given lines is } 96 \text { unit }^{2} \end{aligned}