Question

Area bounded between curves $y^2=4x-4$ and $y=x-1$ is $A.$ Then the value of $\frac{3A}{2}$ is

• A. 04
• B. 4.0
• C. 4.00
• D. 4

Answer 1

Answer: (A), (B), (C), (D)

Intersection points of curves $y^2=4x-4$ and $y=x-1:$
$(x-1{)}^2=4x-4$
$\Rightarrow x^2-6x+5=0$
$\Rightarrow x=1,5$


The area between the curves
$A=\underset{1}{\overset{5}{\int }}[\sqrt{4x-4}-(x-1\left)\right]dx$
$\Rightarrow A={[\frac{4(x-1{)}^{3/2}}{3}-\frac{x^2}{2}+x]}_1^5$
$\Rightarrow A=\frac{32}{3}-\frac{25}{2}+5+\frac{1}{2}-1$
$\Rightarrow A=\frac{8}{3}$
$\therefore \frac{3A}{2}=\frac{3}{2}\times \frac{8}{3}=4$