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)

$\frac{9}{2}$

The graph is shown in the image.

The intersection points of $y=x+1$ and $y=x^2-4x+5$

$x+1=x^2-4x+5\Rightarrow x^2-5x-4=0\Rightarrow x^2-4x-x-4=0$

$x(x-4)-1(x-4)=0\Rightarrow (x-1)(x-4)=0\Rightarrow x=1,4$

$x=1\Rightarrow y=1+1=2$ and $x=4\Rightarrow y=4+1=5$

$\therefore$ The intersection points are $(1,2)$ and $(4,5)$.

Now $A={\int }_1^4[x+1]-{\int }_1^4[x^2-4x+5]$

$A=[\frac{x^2}{2}+x]{|}_1^4-[\frac{x^3}{3}-2x^2+5x]{|}_1^4$

$A=[\frac{16}{2}+4-\frac{1}{2}-1]-[\frac{64}{3}-32+20-\frac{1}{3}+2-5]$

$A=\frac{21}{2}-6$

$A=\frac{9}{2}$ sq.unit