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Mathematics · 2022 · 4 marks
CBSE 2022 · Region 2 · Set 1 · Q12
Evaluate: ∫_0^1 x(1-x)^n d x
Evaluate : \[\int_{0}^{1} x(1-x)^{n} d x \]
Marking-scheme solution
\[\begin{aligned}
I & =\int_{0}^{1} x(1-x)^{n} d x \\
& =\int_{0}^{1}(1-x)[1-(1-x)]^{n} d x \quad \text { [using property] } \\
& =\int_{0}^{1} x^{n}(1-x) d x \\
= & \int_{0}^{1} x^{n} d x-\int_{0}^{1} x^{n+1} d x \\
= & {\left[\frac{x^{n+1}}{n+1}\right]_{0}^{1}-\left[\frac{x^{n+2}}{n+2}\right]_{0}^{1} } \\
& =\frac{1}{n+1}-\frac{1}{n+2} \text { Or } \frac{1}{(n+1)(n+2)}
\end{aligned}
\]
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CBSE Class 12 Mathematics past-paper question from the 2022 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.