
Using integration, find the area of the region bounded by the line $2y=5x+7$, the x – axis, and the lines x = 2 and x = 8.
Hint: We will first draw a rough diagram that represents the area to be calculated. Then we will use the integration to find the area of the region bounded by the line $2y=5x+7$ and the lines x = 2 and x = 8.
Complete step-by-step answer:
Now, before starting the solution we first to understand that the physical significance of \[\int\limits_{a}^{b}{f\left( x \right)dx}\] is,
Now, we can see from the graph that \[\int\limits_{a}^{b}{f\left( x \right)dx}\] is nothing but the area bounded by the line $y=f\left( x \right),x=a,x=b\ and\ x-axis$.
Now, we have to find the area bounded by the curve $2y=5x+7$, the x – axis and the lines x = 2 and x = 8. So, we have,
So, we have the area of the bounded region as $\int\limits_{x=2}^{x=8}{\left( \dfrac{5x+7}{2} \right)dx}$.
We have taken the limits of the integral as x = 2 to x = 8 from graph.
$\Rightarrow \int\limits_{x=2}^{x=8}{\left( \dfrac{5x+7}{2} \right)dx}$
Now, we know that $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$.
$\begin{align}
& \Rightarrow \int\limits_{x=2}^{x=8}{\left( \dfrac{5x}{2}+\dfrac{7}{2} \right)dx} \\
& \Rightarrow \left( \dfrac{5}{2}\dfrac{{{x}^{2}}}{2}+\dfrac{7}{2}x \right)_{2}^{8} \\
& \Rightarrow \left( \dfrac{5}{4}\left( 64-4 \right)+\dfrac{7}{2}\left( 8-2 \right) \right) \\
& \Rightarrow \left( \dfrac{5}{4}\times 60+\dfrac{7}{2}\times 6 \right) \\
& \Rightarrow \left( 5\times 15+7\times 3 \right) \\
& \Rightarrow \left( 75+21 \right) \\
& \Rightarrow 96sq\ units \\
\end{align}$
Therefore, the area bounded by the respective curve is $96sq\ units$.
Note: It is important to note that we have put the limits as x = 2 to x = 8 after referring to the graph of the situation. So, it is important to draw a neat graph depicting all the curves and their point of intersection.












