Mathematics - Area Under Curves Question with Solution | TestHub

Given equation of parabola is $$y=x^2-4x+5$$ $\Rightarrow y=(x-2{)}^2+1$ $\Rightarrow (x-2{)}^2=(y-1)$ and equation of line is $$\begin{array}{r}y=x+1\\ \Rightarrow x-y=-1\end{array}$$On putting the value of $(y-1)$ from Eqs. (ii) in (1). we get $$(x-2{)}^2=x\Rightarrow x^2+4-4x=x$$ $\Rightarrow x^2-5x+4=0$ $\Rightarrow x^2-4x-x+4=0$ $\Rightarrow x(x-4)-1(x-4)=0$ $\Rightarrow (x-1)(x-4)=0$ or $x=1,4$ then from is (ii) $y=2,5$ $\therefore$ Required area $={\int }_1^4\left\{(x+1)-\left(x^2-4x+5\right)\right\}dx$ $={\int }_1^4\left(-x^2+5x-4\right)dx={\left[\frac{-x^3}{3}+\frac{5x^2}{2}-4x\right]}_1^4$ $=\left[-\frac{64}{3}+40-16+\frac{1}{3}-\frac{5}{2}+4\right]$ $=\left(-21-\frac{3}{2}+28\right)=\frac{9}{2}$