If the area (in square units) bounded by the curves \( y^2=4x \) and \( y=x \) is \( \displaystyle\frac{k^3}{3} \), then the value of k is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
The curves meet where \( x^2=4x \), that is, at x = 0 and x = 4, the points (0, 0) and (4, 4). Between them the parabola \( y=2\sqrt{x} \) lies above the line y = x.
\( \displaystyle \text{Area}=\int_{0}^{4}\left(2\sqrt{x}-x\right)dx \)
\( \displaystyle =\left[\frac{4}{3}x^{3/2}-\frac{x^2}{2}\right]_{0}^{4} \)
\( \displaystyle =\frac{4}{3}(8)-\frac{16}{2} \)
\( \displaystyle =\frac{32}{3}-8=\frac{8}{3} \)
So
\( \displaystyle \frac{k^3}{3}=\frac{8}{3} \)
\( k^3=8 \)
\( k=2 \)
Tip: Worth remembering: the area between \( y^2=4ax \) and \( y=mx \) is \( \frac{8a^2}{3m^3} \). With a = 1 and m = 1 it gives \( \frac{8}{3} \) at once.