Question

The area bounded by the parabola \( y = x^2 + 4 \) and the straight line passing through the points \( (-1,2) \quad {and} \quad (1,6) \) is (in square units)

• A. \( \frac{20}{3} \)
• B. \( \frac{4}{3} \)
• C. \( \frac{8}{3} \)
• D. \( \frac{16}{3} \)
• E. \( \frac{14}{3} \)

Hint:

To find the bounded area between a curve and a line, compute \( \int (y_{{upper}} - y_{{lower}})dx \).

Answer 1

The Correct Option is B

First, find the equation of the straight line passing through the given points: \[ m = \frac{6 - 2}{1 - (-1)} = \frac{4}{2} = 2 \] Equation of the line: \[ y - 2 = 2(x + 1) \] \[ y = 2x + 4 \] Now, find the points of intersection by solving: \[ x^2 + 4 = 2x + 4 \] \[ x^2 - 2x = 0 \] \[ x(x - 2) = 0 \] \[ x = 0, 2 \] The required area is: \[ A = \int_0^2 [(2x + 4) - (x^2 + 4)] dx \] \[ A = \int_0^2 (2x - x^2) dx \] Evaluating: \[ A = \left[ x^2 - \frac{x^3}{3} \right]_0^2 \] \[ = \left[ 4 - \frac{8}{3} \right] - [0 - 0] \] \[ = \frac{12}{3} - \frac{8}{3} = \frac{4}{3} \] Thus, the correct answer is (B).