MHT CET · Maths · Definite Integration
\(\int_0^1 x(1-x)^n d x=\)
- A \(\frac{n+3}{(n+1)(n+2)}\)
- B \(\frac{1}{(n+1)(n+2)}\)
- C \(\frac{2 n+3}{(n+1)(n+2)}\)
- D \(\frac{4}{(n+1)(n+2)}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{(n+1)(n+2)}\)
Step-by-step Solution
Detailed explanation
\(\int_0^1 x(1-x)^n d x=\int_0^1(1-x)^n d x[\because \int_0^a f(x) d x=\) \(\int_0^a f(a-x) d x] \)
\( =\int_0^1\left(x^n-x^{n+1}\right) d x=\left[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right]_0^1 \)
\( =\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)}\)
\( =\int_0^1\left(x^n-x^{n+1}\right) d x=\left[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right]_0^1 \)
\( =\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)}\)
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