JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Area Under Curves

The area (in sq. units) in the first quadrant bounded by the parabola

y = x^{2} + 1

, the tangent to it at the point

(2, 5)

and the coordinate axes is:

Answer: (B)

Solution

The parabola is

y = x^{2} + 1

, and at

(2, 5)

its slope is

2 x = 4

, so the tangent is

y = 4 x - 3

, which cuts the

x

-axis at

x = \frac{3}{4}

. The required area is the region under the parabola from

0

2

less the triangle beneath the tangent:

\int_{0}^{2} (x^{2} + 1) d x - \frac{1}{2} (2 - \frac{3}{4}) (5) = (\frac{8}{3} + 2) - \frac{1}{2} \cdot \frac{5}{4} \cdot 5 = \frac{14}{3} - \frac{25}{8} = \frac{112 - 75}{24} = \frac{37}{24}

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option