Question

What is the area bounded by the curve $y\le x^2+3x,0\le y\le 4,0\le x\le 3$

• A. $\frac{59}{6}$
• B. $\frac{57}{4}$
• C. $\frac{59}{3}$
• D. $\frac{57}{6}$

Answer 1

The correct option is A

$\frac{59}{6}$

Explanation of correct answer :

Finding area within the curves :

The graph of the curves is as shown:

The shaded area is the area within the curves.

To get the point of intersection, equate curves of the equation $y=x^2+3x\text{and}y=4$

$$\begin{gathered} \Rightarrow x^2+3x=4 \\ \Rightarrow x^2+3x-4=0 \\ \Rightarrow x^2+4x-x-4=0 \\ \Rightarrow \left(x+4\right)(x-1)=0 \\ \Rightarrow x=1or-4 \end{gathered}$$

Required area $={\int }_0^1\left(x^2+3x\right)dx+\text{area of rectangle}$

$$\begin{gathered} ={\left[\frac{x^3}{3}+\frac{3x^2}{2}\right]}_0^1+2\times 4 \\ =\frac{1}{3}+\frac{3}{2}+8 \\ =\frac{59}{6} \end{gathered}$$

Hence, the area within the curve is $\frac{59}{6}$ sq. units.

Thus, the correct answer is option(A).