The area of the region bounded by the curve x^2=4y , the line x=4y-2…

The correct answer is Option (3) → \( \displaystyle\frac{5}{24} \) Square units

Points where the boundaries meet:

    \( \displaystyle \text{curve and x-axis: } (0,0) \)
    \( \displaystyle \text{line and x-axis: } (-2,0) \)

Curve and line: \( y=\frac{x+2}{4} \), so

    \( \displaystyle x^2=x+2\ \Rightarrow\ (x-2)(x+1)=0 \)

    \( \displaystyle x=-1\ \text{or}\ x=2 \)

The region with all three as boundaries has corners (−2, 0), (−1, 1/4) and (0, 0). Its upper boundary is the line from x = −2 to x = −1 and the parabola from x = −1 to x = 0, with the x-axis below:

    \( \displaystyle \text{Area}=\int_{-2}^{-1}\frac{x+2}{4}\,dx+\int_{-1}^{0}\frac{x^2}{4}\,dx \)

    \( \displaystyle =\left[\frac{(x+2)^2}{8}\right]_{-2}^{-1}+\left[\frac{x^3}{12}\right]_{-1}^{0} \)

    \( \displaystyle =\frac{1}{8}+\frac{1}{12} \)

    \( \displaystyle =\frac{5}{24} \)

The area is \( \frac{5}{24} \) square units.

Why the other options are incorrect

• Option (1) \( \displaystyle\frac{9}{8} \) Square units — Incorrect. \( \frac{9}{8} \) is the area enclosed between the line and the parabola alone, from x = −1 to x = 2; that region is not bounded by the x-axis.

• Option (2) \( \displaystyle\frac{9}{24} \) Square units — Incorrect. The area is \( \frac{1}{8}+\frac{1}{12}=\frac{5}{24} \), not \( \frac{9}{24} \).

• Option (4) \( \displaystyle\frac{11}{24} \) Square units — Incorrect. The area is \( \frac{5}{24} \), not \( \frac{11}{24} \).