Question

The area (in sq units) bounded by the curves$y=3x$ and $y=x^2$ is

• A. $10$
• B. $5$
• C. $4.5$
• D. $9$

Answer 1

The correct option is C

$4.5$

Explanation for Correct Answer:

Determining the area bounded by the curve.

Given, the curves are $y=3x$ and $y=x^2$

Substituting equation $y=3x$ in equation $y=x^2$, we have

$$\begin{gathered} y=x^2 \\ \Rightarrow 3x=x^2 \\ \Rightarrow x^2-3x=0 \\ \Rightarrow x\left(x-3\right)=0 \\ \Rightarrow x=0\text{or}x=3 \end{gathered}$$

These are the points of intersection of the given curves

Now the area of the bounded region is given by

$\begin{array}{l}A={\int }_0^3\left(y_{uppercurve}-y_{lowercurve}\right)dx\\ A={\int }_0^3\left(3x-x^2\right)dx\\ ={\left[3\times \frac{x^2}{2}-\frac{x^3}{3}\right]}_0^3\\ =\left[\left(3\times \frac{3^2}{2}-\frac{3^3}{3}\right)-0\right]\\ =\left[\left(\frac{27}{2}-\frac{27}{3}\right)\right]\\ =27\times \frac{3-2}{6}\\ =\frac{27}{6}\end{array}$

Therefore the area bounded by the given curve is equal to $\frac{27}{6}$ or $4.5\text{square units}$

Hence, the correct answer is Option (C)