A man plans to jump and climb the stairs. The path (as shown in the figure) he moves for the first jump is given by \(y = y(x) = x + \sqrt{2x - 3x^2}\) where ‘x’ (in meters) is the horizontal distance moved and ‘y’ (in meters) is the corresponding height. Assuming that his initial point is at origin.
Based on the above information answer the Questions
Evaluate \(\int_{0}^{\frac{1}{3}}\displaystyle y\,dx\)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{18} + \sqrt{3}\,\frac{\pi}{36}\)
Split the integral:
\( \displaystyle \int_0^{1/3} y\,dx = \int_0^{1/3} x\,dx + \int_0^{1/3}\sqrt{2x - 3x^2}\,dx \)
The first part:
\( \displaystyle \int_0^{1/3} x\,dx = \frac{1}{2}\cdot\frac{1}{9} = \frac{1}{18} \)
For the second part, complete the square:
\( \displaystyle 2x - 3x^2 = 3\left(\frac{2x}{3} - x^2\right) = 3\left[\left(\frac{1}{3}\right)^2 - \left(x - \frac{1}{3}\right)^2\right] \)
Put \(t = x - \frac{1}{3}\); then t runs from \(-\frac{1}{3}\) to 0, and with \(a = \frac{1}{3}\):
\( \displaystyle \int_0^{1/3}\sqrt{2x - 3x^2}\,dx = \sqrt{3}\int_{-1/3}^{0}\sqrt{a^2 - t^2}\,dt \)
Using \(\int\sqrt{a^2 - t^2}\,dt = \frac{t}{2}\sqrt{a^2 - t^2} + \frac{a^2}{2}\sin^{-1}\frac{t}{a}\):
\( \displaystyle \int_{-1/3}^{0}\sqrt{a^2 - t^2}\,dt = 0 - \left[0 + \frac{1}{18}\sin^{-1}(-1)\right] \)
\( \displaystyle = \frac{1}{18}\cdot\frac{\pi}{2} = \frac{\pi}{36} \)
(This is a quarter of the circle of radius \(\frac{1}{3}\).) So
\( \displaystyle \int_0^{1/3} y\,dx = \frac{1}{18} + \sqrt{3}\,\frac{\pi}{36} \)