Question

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 :

• A. \(\frac{8}{3}\)
• B. \(\frac{14}{3}\)
• C. \(\frac{187}{24}\)
• D. \(\frac{37}{24}\)

Hint:

Find the tangent y = 4x - 3 at (2,5); the area is the area under the parabola from 0 to 2 minus the triangle under the tangent between x = 3/4 and x = 2.

Answer 1

The Correct Option is D

\(\frac{dy}{dx} = 2x = 4\) at x = 2, so the tangent at (2, 5) is \(y - 5 = 4(x-2)\), i.e. \(y = 4x - 3\). It cuts the x-axis at \(x = \frac34\).
In the first quadrant the region is bounded above by the parabola, on the left by the y-axis, below by the x-axis (for \(0 \le x \le \frac34\)) and by the tangent (for \(\frac34 \le x \le 2\)).
Area under the parabola from 0 to 2: \(\int_0^2 (x^2+1)\,dx = \frac83 + 2 = \frac{14}{3}\).
Area under the tangent from \(\frac34\) to 2 (a triangle with base \(\frac54\) and height 5): \(\frac12\cdot\frac54\cdot 5 = \frac{25}{8}\).
Required area = \(\frac{14}{3} - \frac{25}{8} = \frac{112 - 75}{24} = \frac{37}{24}\).
Answer: (4) \(\frac{37}{24}\)