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)

$y=x^2+1\dots (1)$
$\frac{dy}{dx}=2x_1$
$m_N=-\frac{1}{2x_1}=-1\Rightarrow x_1=\frac{1}{2}$
$y_1=1+\frac{1}{4}=\frac{5}{4}$
equation of normal
$y-\frac{5}{4}=-1\left(x-\frac{1}{2}\right)$
$y+x=\frac{1}{2}+\frac{5}{4}=\frac{7}{4}$
$\text{ put }y=\frac{7}{4}-x\text{ in (1) }$
$\frac{7}{4}-x=x^2+1$

$x^2+x-\frac{3}{4}=0\Rightarrow x_1{x}_2=-\frac{3}{4};x_2\cdot \frac{1}{2}=-\frac{3}{4}\Rightarrow x_2=-\frac{3}{2}$
${A=\underset{-3/2}{\overset{1/2}{\int }}\left[\left(\frac{7}{4}-x\right)-\left(x^2+1\right)\right]dx=\underset{-3/2}{\overset{1/2}{\int }}\left(\frac{3}{4}-x-x^2\right)dx=\frac{3x}{4}-\frac{x^2}{2}-\frac{x^3}{3}]}_{-3/2}^{1/2}$
$=\left(\frac{3}{2}\cdot \frac{1}{2}-\frac{1}{8}-\frac{1}{24}\right)-\left(-\frac{9}{8}-\frac{9}{8}+\frac{27}{24}\right)=\frac{2}{8}-\frac{1}{24}+\frac{9}{8}=\frac{5}{24}+\frac{27}{24}=\frac{32}{24}=\frac{4}{3}=\frac{16}{12}$