Question

The area bounded by the curves $y=\frac{1}{4}|4-x^2|$ and $y=7-|x|$ is

• A. $18$
• B. $32$
• C. $36$
• D. $64$

Answer 1

Answer: (B)

$32$

The graphs can be drawn as shown in the figure.

We need to find the area of the shaded portion. Since the area is symmetric about the y-axis, we will find the area on the right-hand side of the y-axis and later, multiply it by 2.

First, we find the area bounded by the straight line and the x-axis. Then, we subtract the area bounded by the parabola and the x-axis from our previous result.

Now, point A is the intersection point of both the graphs for $x>0$. So, we have

$\frac{x^2}{4}-1=7-x$

$⟹x^2-4=28-4x$

$⟹x^2+4x-32=0⟹(x+8)(x-4)=0$

$⟹x=4$ (since $x>0$)

$⟹y=7-x=7-4=3$

So, $A\equiv (4,3)$

Area bounded by the straight line and the x-axis $={\int }_0^4(7-x)dx={(7x-\frac{x^2}{2})}_0^4=28-\frac{4^2}{2}=20$

Area bounded by the parabola $={\int }_0^2(1-\frac{x^2}{4})dx+{\int }_2^4(\frac{x^2}{4}-1)dx={(x-\frac{x^3}{12})}_0^2+{(\frac{x^3}{12}-x)}_2^4=4$

So, the shaded area on the right hand side becomes $(20-4)=16$

$\therefore$ The total area $=2\times 16=32$