Home » Blog » Find the integral from 0 to 1 of x4 (1-x)20

Find the integral from 0 to 1 of x4 (1-x)20

by RoRi

Evaluate the following integral:

    \[ \int_0^1 x^4 (1-x)^{20} \, dx. \]

First, we make the substitution

    \[ u = 1-x \qquad \implies \qquad du = - dx. \]

This gives us new bounds of integration

    \[ u(0) = 1 \qquad \text{and} \qquad u(1) = 0. \]

Therefore, we have

    \begin{align*} \int_0^1 x^4 (1-x)^{20} \, dx &= \int_1^0 -(1-u)^4 u^{20} \, du \\ &= \int_0^1 (1-u)^4 u^{20} \, du. \end{align*}

(This still isn’t great, computationally, but we would rather take the fourth power of (1-u) than the 20th power of (1-x).) So, multiplying out the integrand we have

    \begin{align*} (1-u)^4 u^{20} &= u^{20} ( 1 - 4u + 6u^2 - 4u^3 + u^4) \\ &= u^{24} - 4u^{23} + 6u^{22} - 4u^{21} + u^{20}. \end{align*}

Therefore,

    \begin{align*} \int_0^1 (1-u)^4 u^{20} \, du &= \int_0^1 \left( u^{24} - 4u^{23} + 6u^{22} - 4u^{21} + u^{20} \right) \, du \\ &= \left( \frac{u^{25}}{25} - \frac{u^{24}}{6} + \frac{6u^{23}}{23} - \frac{2u^{22}}{11} + \frac{u^{21}}{21} \right) \Big \rvert_0^1 \\ &= \frac{1}{25} - \frac{1}{6} + \frac{6}{23} - \frac{2}{11} + \frac{1}{21} \\ &= \frac{1}{265650}. \end{align*}

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):