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

• A

  $\frac{1}{2}$
• B

  2
• C

  3
• Correct answer
  D

  $\frac{9}{2}$

Solution

Given equation of parabola is $$ y=x^{2}-4 x+5 $$ $\Rightarrow \quad y=(x-2)^{2}+1$ $\Rightarrow \quad(x-2)^{2}=(y-1)$ and equation of line is $$ \begin{array}{r} y=x+1 \\ \Rightarrow \quad x-y=-1 \end{array} $$ On putting the value of $(y-1)$ from Eqs. (ii) in (1). we get $$ (x-2)^{2}=x \Rightarrow x^{2}+4-4 x=x $$ $\Rightarrow \quad x^{2}-5 x+4=0$ $\Rightarrow x^{2}-4 x-x+4=0$ $\Rightarrow \quad x(x-4)-1(x-4)=0$ $\Rightarrow \quad(x-1)(x-4)=0$ or $x=1,4$ then from is (ii) $y=2,5$ $\therefore$ Required area $=\int_{1}^{4}\left\{(x+1)-\left(x^{2}-4 x+5\right)\right\} d x$ $=\int_{1}^{4}\left(-x^{2}+5 x-4\right) d x=\left[\frac{-x^{3}}{3}+\frac{5 x^{2}}{2}-4 x\right]_{1}^{4}$ $=\left[-\frac{64}{3}+40-16+\frac{1}{3}-\frac{5}{2}+4\right]$ $=\left(-21-\frac{3}{2}+28\right)=\frac{9}{2}$

Hence, the correct answer is option: 'D'.