Question

Area enclosed by the curve $y=x^2+1$ and a normal drawn to it with gradient -1 , is equal to

• A. $\frac{8}{12}$
• B. $\frac{16}{12}$
• C. $\frac{19}{12}$
• D. $\frac{43}{12}$

Answer 1

Answer: (B)

$ y=x^2+1 \ldots(1) $
$\frac{d y}{d x}=2 x_1$
$m_N=-\frac{1}{2 x_1}=-1 \Rightarrow x_1=\frac{1}{2}$
$y _1=1+\frac{1}{4}=\frac{5}{4}$
equation of normal
$y-\frac{5}{4}=-1\left(x-\frac{1}{2}\right) $
$y+x=\frac{1}{2}+\frac{5}{4}=\frac{7}{4} $
$\text { put } y=\frac{7}{4}-x \text { in (1) } $
$\frac{7}{4}-x=x^2+1$

$x^2+x-\frac{3}{4}=0 \Rightarrow x_1 x_2=-\frac{3}{4} ; x_2 \cdot \frac{1}{2}=-\frac{3}{4} \Rightarrow x_2=-\frac{3}{2} $
$\left.A=\int\limits_{-3 / 2}^{1 / 2}\left[\left(\frac{7}{4}-x\right)-\left(x^2+1\right)\right] d x=\int\limits_{-3 / 2}^{1 / 2}\left(\frac{3}{4}-x-x^2\right) d x=\frac{3 x}{4}-\frac{x^2}{2}-\frac{x^3}{3}\right]_{-3 / 2}^{1 / 2}$
$=\left(\frac{3}{2} \cdot \frac{1}{2}-\frac{1}{8}-\frac{1}{24}\right)-\left(-\frac{9}{8}-\frac{9}{8}+\frac{27}{24}\right)=\frac{2}{8}-\frac{1}{24}+\frac{9}{8}=\frac{5}{24}+\frac{27}{24}=\frac{32}{24}=\frac{4}{3}=\frac{16}{12}$