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

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*}

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