Question

The area of the region bounded by the parabola y = x² + 1 and the straight line x + y = 3 is given by
(a) $\frac{45}{7}$

(b) $\frac{25}{4}$

(c) $\frac{\mathrm{\pi }}{18}$

(d) $\frac{9}{2}$

Answer 1

$\left(\mathrm{d}\right)\frac{9}{2}$



To find the point of intersection of the parabola y = x² + 1 and the line x + y = 3 substitute y = 3 − x in y = x² + 1
$$\begin{gathered} 3-x=x^2+1 \\ \Rightarrow x^2+x-2=0 \\ \Rightarrow \left(x-1\right)\left(x+2\right)=0 \\ \Rightarrow x=1\mathrm{or}x=-2 \\ \therefore y=2\mathrm{or}y=5 \end{gathered}$$
So, we get the points of intersection A(−2, 5) and C(1, 2).
Therefore, the required area ABC,
$$\begin{gathered} A={\int }_{-2}^1\left(y_1-y_2\right)\mathit{d}x\mathit{}\left(\mathrm{Where},y_1=3-x\mathrm{and}{y}_2=x^2+1\right) \\ ={\int }_{-2}^1\left[\left(3-x\right)-\left(x^2+1\right)\right]\mathit{d}x \\ ={\int }_{-2}^1\left(3-x-x^2-1\right)\mathit{d}x \\ ={\int }_{-2}^1\left(2-x-x^2\right)\mathit{d}x \\ ={\left[2x-\frac{x^2}{2}-\frac{x^3}{3}\right]}_{-2}^1 \\ =\left[2\left(1\right)-\frac{{\left(1\right)}^2}{2}-\frac{{\left(1\right)}^3}{3}\right]-\left[2\left(-2\right)-\frac{{\left(-2\right)}^2}{2}-\frac{{\left(-2\right)}^3}{3}\right] \\ =\left[2-\frac{1}{2}-\frac{1}{3}\right]-\left[-4-2+\frac{8}{3}\right] \\ =2-\frac{1}{2}-\frac{1}{3}+4+2-\frac{8}{3} \\ =8-\frac{1}{2}-\frac{9}{3} \\ =5-\frac{1}{2} \\ =\frac{9}{2}\mathrm{square}\mathrm{units} \end{gathered}$$