Question

The area of the region bounded by the parabola $y=x^2-4x+5$ and the straight line $y=x+1$ is

• A. $\frac{1}{2}$
• B. $2$
• C. $3$
• D. $\frac{9}{2}$

Answer 1

Answer: (D)

$\frac{9}{2}$

Given equation of parabola is
$y=x^2-4x+5$
$\Rightarrow$ $y={(x-2)}^2+1$
$\Rightarrow$ ${(x-2)}^2=(y-1)....\left(i\right)$
and equation of line is
$y=x+1$
$\Rightarrow$ $x-y=-\mathrm{1.....}\left)\right(ii)$

On putting the value of $(y-1)$ from eqs $\left(ii\right)$ in $\left(i\right)$ we get
${(x-2)}^2=x$ $\Rightarrow$ $x^2+4-4x=x$
$\Rightarrow$ $x^2-5x+4=0$
$\Rightarrow$ $x^2-4x-x+4=0$
$\Rightarrow$ $x(x-4)-1(x-4)=0$
$\Rightarrow$ $(x-1)(x-4)=0$ or $x=1,4$
then from is $\left(ii\right)$ $y=2,5$

$\therefore$ Required area $={\int }_1^4(-x^2+5x-4)dx={[\frac{-x^3}{3}+\frac{5x^2}{2}-4x]}_1^4$
$=[-\frac{64}{3}+40-16+\frac{1}{3}-\frac{5}{2}+4]$

$=(-21-\frac{5}{2}+28)=\frac{9}{2}$