<s>[INST] How to find $\int_0^1 x^4(1-x)^5dx$ quickly? This question came in the Rajshahi University admission exam 2018-19
Q) $\int_0^1 x^4(1-x)^5dx$=?
(a) $\frac{1}{1260}$
(b) $\frac{1}{280}$
(c)$\frac{1}{315}$
(d) None
This is a big integral (click on show steps):
$$\left[-\dfrac{\left(x-1\right)^6\left(126x^4+56x^3+21x^2+6x+1\right)}{1260}\right]_0^1=\frac{1}{1260}$$
It takes a lot of time to compute. How can I compute this quickly (30 seconds) using a shortcut?
 [/INST] If you don't have the beta function formula memorized, you can use integration by parts repeatedly
\begin{align*}
\int_{0}^{1}x^4(1-x)^5\,dx &= \left[x^4 \cdot -\dfrac{1}{6}(1-x)^6\right]_{0}^{1} - \int_{0}^{1}4x^3 \cdot -\dfrac{1}{6}(1-x)^6\,dx
\\
&= \dfrac{4}{6}\int_{0}^{1}x^3(1-x)^6\,dx
\\
&= \dfrac{4}{6}\left[x^3 \cdot -\dfrac{1}{7}(1-x)^7\right]_{0}^{1} - \dfrac{4}{6}\int_{0}^{1}3x^2 \cdot -\dfrac{1}{7}(1-x)^7\,dx
\\
&= \dfrac{4 \cdot 3}{6 \cdot 7}\int_{0}^{1}x^2(1-x)^7\,dx
\\
&= \dfrac{4 \cdot 3}{6 \cdot 7}\left[x^2 \cdot -\dfrac{1}{8}(1-x)^8\right]_{0}^{1} - \dfrac{4 \cdot 3}{6 \cdot 7}\int_{0}^{1}2x \cdot -\dfrac{1}{8}(1-x)^8\,dx
\\
&= \dfrac{4 \cdot 3 \cdot 2}{6 \cdot 7 \cdot 8}\int_{0}^{1}x(1-x)^8\,dx
\\
&= \dfrac{4 \cdot 3 \cdot 2}{6 \cdot 7 \cdot 8}\left[x \cdot -\dfrac{1}{9}(1-x)^9\right]_{0}^{1} - \dfrac{4 \cdot 3 \cdot 2}{6 \cdot 7 \cdot 8}\int_{0}^{1}1 \cdot -\dfrac{1}{9}(1-x)^9\,dx
\\
&= \dfrac{4 \cdot 3 \cdot 2}{6 \cdot 7 \cdot 8 \cdot 9}\int_{0}^{1}(1-x)^9\,dx
\\
&= \dfrac{4 \cdot 3 \cdot 2}{6 \cdot 7 \cdot 8 \cdot 9}\left[-\dfrac{1}{10}(1-x)^{10}\right]_{0}^{1}
\\
&= \dfrac{4 \cdot 3 \cdot 2}{6 \cdot 7 \cdot 8 \cdot 9 \cdot 10}
\\
&= \dfrac{1}{1260}
\end{align*}
This looks fairly long since I wrote out all the steps in detail. However, if you notice the pattern, you can probably arrive at the answer faster. Also, if are familiar with the Tabular Integration Method, then you only need to do a fairly short amount of scratchwork.
</s>