∫_0^2 x (2-x)^n dx is equal to

The correct answer is Option (3) → \(\displaystyle\frac{2^{n+2}}{(n+1)(n+2)}\)

Let I be the integral. By the property \(\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx\) with a = 2, x and 2 − x change places, which moves the power onto a single x:

    \( \displaystyle I = \int_{0}^{2} (2 - x)\,x^n\,dx \)

    \( \displaystyle I = \int_{0}^{2} \left(2x^n - x^{n+1}\right) dx \)

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

    \( \displaystyle I = \frac{2^{n+2}}{n+1} - \frac{2^{n+2}}{n+2} \)

    \( \displaystyle I = 2^{n+2} \cdot \frac{(n+2) - (n+1)}{(n+1)(n+2)} \)

    \( \displaystyle I = \frac{2^{n+2}}{(n+1)(n+2)} \)

Shortcut: Put n = 0. The integral becomes \(\int_{0}^{2} x\,dx = 2\). Only Option (3) gives 2 at n = 0 (the others give \(\frac{4}{6}\), \(\frac{8}{3}\) and 1).