Question

The integral \(\int_0^1x(1-x)^n dx\) is equal to :

• A. \(\frac{1}{(n+2)(n+3)}\)
• B. \(\frac{1}{(n+1)(n+2)}\)
• C. \(\frac{1}{n(n+1)}\)
• D. \(\frac{1}{(n-1)(n-2)}\)

Answer 1

The Correct Option is B

To solve the integral \(\int_0^1 x(1-x)^n \, dx\), we use integration by parts. Let \(u = x\) and \(dv = (1-x)^n dx\). Then, \(du = dx\) and we find \(v\) by integrating \(dv\):

\[\begin{aligned} v &= \int (1-x)^n \, dx. \end{aligned}\]

Using the substitution \(w = 1-x\), \(dw = -dx\), the integral becomes:

\[\begin{aligned} v &=-\int w^n \, dw = -\frac{w^{n+1}}{n+1} = -\frac{(1-x)^{n+1}}{n+1}. \end{aligned}\]

Now apply integration by parts:

\[\begin{aligned} \int u \, dv &= uv - \int v \, du \\ &= x\left(-\frac{(1-x)^{n+1}}{n+1}\right) \bigg|_0^1 - \int \left(-\frac{(1-x)^{n+1}}{n+1}\right) \, dx. \end{aligned}\]

Calculate \(uv\) at the bounds from 0 to 1:

\[\begin{aligned} &\left[-\frac{x(1-x)^{n+1}}{n+1}\right]_0^1 = \left[-\frac{1 \cdot 0^{n+1}}{n+1}\right] - \left[-\frac{0 \cdot 1^{n+1}}{n+1}\right] = 0. \end{aligned}\]

Now compute the remaining integral:

\[\begin{aligned} \frac{1}{n+1} \int (1-x)^{n+1} \, dx &= \frac{1}{n+1} \cdot \left[-\frac{(1-x)^{n+2}}{n+2}\right]_0^1 \\ &= \frac{-1}{(n+1)(n+2)} \left[(1-1)^{n+2} - (1-0)^{n+2}\right] \\ &= \frac{-1}{(n+1)(n+2)} (0 - 1) = \frac{1}{(n+1)(n+2)}. \end{aligned}\]

Thus, the integral evaluates to \(\frac{1}{(n+1)(n+2)}\), which matches the correct option.