The area of the region bounded by the curve \(x^2 = 4y\) and the straight line \(x = 4y - 2\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{9}{8}\) sq. unit
The line is \(y = \frac{x+2}{4}\) and the parabola is \(y = \frac{x^2}{4}\). At the points of intersection:
\(x^2 = x + 2\)
\(x^2 - x - 2 = 0\)
\((x - 2)(x + 1) = 0,\quad x = -1 \text{ or } 2\)
Between x = −1 and x = 2 the line lies above the parabola (at x = 0 the line gives \(\frac{1}{2}\) and the parabola 0). So
\( \displaystyle \text{Area} = \int_{-1}^{2}\left(\frac{x+2}{4} - \frac{x^2}{4}\right)dx \)
\( \displaystyle = \frac{1}{4}\left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2} \)
\( \displaystyle = \frac{1}{4}\left[\left(2 + 4 - \frac{8}{3}\right) - \left(\frac{1}{2} - 2 + \frac{1}{3}\right)\right] \)
\( \displaystyle = \frac{1}{4}\left[\frac{10}{3} + \frac{7}{6}\right] \)
\( \displaystyle = \frac{1}{4} \times \frac{27}{6} = \frac{9}{8} \)
NCERT Exemplar:
Application of Integrals
