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The area of the region bounded by the curve x2=4yx^2 = 4y and the straight line x=4y−2x = 4y - 2 is:

Solution

✅ Correct Option: 4

From the parabola: y=x2/4y = x^2/4. From the line: y=(x+2)/4y = (x+2)/4.

Intersection: x2=x+2⇒x=−1,2x^2 = x + 2 \Rightarrow x = -1, 2.

Area =∫−12[x+24−x24]dx=14[x22+2x−x33]−12= \int_{-1}^{2}\left[\frac{x+2}{4} - \frac{x^2}{4}\right]dx = \frac{1}{4}\left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2}

=14[103+76]=14⋅276=98= \frac{1}{4}\left[\frac{10}{3} + \frac{7}{6}\right] = \frac{1}{4} \cdot \frac{27}{6} = \frac{9}{8} sq. unit.

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