Question

The area enclosed by $y=3x-5,y=0,x=3$ and $x=5$ is

• A. $12\mathrm{sq}\mathrm{units}$
• B. $13\mathrm{sq}\mathrm{units}$
• C. $\frac{131}{2}\mathrm{sq}\mathrm{units}$
• D. $14\mathrm{sq}\mathrm{units}$

Answer 1

The correct option is D

$14\mathrm{sq}\mathrm{units}$

Explanation for correct option

Given, $y=3x-5,y=0,x=3$ and $x=5$

Step 1: Finding Intersection points of the curves:

line $x=3$ intersects $y=3x-5$ at

$$\begin{gathered} y=3x-5 \\ y=3\left(3\right)-5 \\ =4 \end{gathered}$$

Point of intersection$=\left(3,4\right)$

Line $x=5$ intersects $y=3x-5$ at

$$\begin{gathered} y=3x-5 \\ y=3\left(5\right)-5 \\ =10 \end{gathered}$$

Point of intersection$=\left(5,10\right)$

Step 2: Finding the area of the region bounded by the curves:

The area enclosed by these curves is,

$$\begin{gathered} \mathrm{Area}={\int }_3^5\left(3\mathrm{x}-5\right)d\mathrm{x} \\ ={\left[\frac{3}{2}{\mathrm{x}}^2-5\mathrm{x}\right]}_3^5 \\ =\left[\frac{3}{2}\left(5^2-3^2\right)-5\left(5-3\right)\right] \\ =\left[24-10\right] \\ =14\mathrm{square}\mathrm{units} \end{gathered}$$

Therefore the area enclosed by the given curves is equal to $14\mathrm{square}\mathrm{units}$.

Hence, option (D) is the correct answer.