The area bounded by the curve x^2 = 4y and the line x = 4y - 2 is:

The correct answer is Option (1) → \(\displaystyle\frac{9}{8}\)

In terms of x, the two curves are:

    \( \displaystyle \text{parabola: } y = \frac{x^2}{4}, \qquad \text{line: } y = \frac{x + 2}{4} \)

Points of intersection: put \(4y = x^2\) in the line:

    \(x = x^2 - 2\)
    \(x^2 - x - 2 = 0\)
    \((x - 2)(x + 1) = 0,\quad x = -1 \text{ or } 2\)

Between x = −1 and x = 2 the line is above the parabola (at x = 0: \(\frac{2}{4} \gt 0\)). So the area is

    \( \displaystyle A = \int_{-1}^{2}\left(\frac{x + 2}{4} - \frac{x^2}{4}\right)dx \)

    \( \displaystyle = \frac{1}{4}\left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2} \)

    \( \displaystyle = \frac{1}{4}\left[\left(2 + 4 - \frac{8}{3}\right) - \left(\frac{1}{2} - 2 + \frac{1}{3}\right)\right] \)

    \( \displaystyle = \frac{1}{4}\left[\frac{10}{3} + \frac{7}{6}\right] \)

    \( \displaystyle = \frac{1}{4} \times \frac{9}{2} = \frac{9}{8} \)

NCERT Exemplar:
Application of Integrals