Question bank/JEE Main/Mathematics

The area (in sq. units) bounded by the parabola y = x^2 -1,
the tangent at the point (2, 3) to it and the y axis is

A$\frac{56}{3}$B$\frac{8}{3}$C$\frac{32}{3}$D$\frac{14}{3}$

Show answer and solution▾

tangentat (2,3) is $\frac{y + 3}{2} = 2x - 1$
y = 4x-5
[IMAGE] area $= \int_{- 5}^{3}\mspace{2mu} xdy - \int_{- 1}^{3}\mspace{2mu} xdy$
$= \int_{- 5}^{3}\mspace{2mu}\left( \frac{y + 5}{4} \right) - \int_{- 1}^{3}\mspace{2mu}\sqrt{y + 1}dy$
$= |\left. \ \frac{\frac{y^{2}}{2} + 5y}{4} \right|_{- 5}^{3} - \left| \frac{2}{3}(y + 1)^{3/2} \right|_{- 1}^{3}$
$= |\frac{\left( \frac{9}{2} + 15 \right) - \left( \frac{25}{2} - 25 \right)}{4}| = | - \frac{16}{3}| = \frac{8}{3}$

Source: Grass-G / M-28 Mathematics Paper FST 16 Jan.docx