Skip to content
Question

Q.(b) Evaluate : ∫ 𝒙(𝟏 βˆ’ 𝒙)𝒏𝒅𝒙; (𝐰𝐑𝐞𝐫𝐞 𝒏 ∈ 𝑡). 𝟏 𝟎

CBSESample paperShortΒ· 3mImportanceβ˜…β˜…β˜…β˜…β˜…est
βœ“ Free question

Substituting t=1βˆ’xt = 1 - x turns the integral into ∫01(tnβˆ’tn+1) dt\int_0^1 (t^n - t^{n+1})\,dt, giving ∫01x(1βˆ’x)n dx=1(n+1)(n+2)\displaystyle \int_0^1 x(1-x)^n\,dx = \frac{1}{(n+1)(n+2)}.

Evaluate ∫01x(1βˆ’x)n dx\displaystyle \int_0^1 x(1-x)^n\,dx for n∈Nn \in \mathbb{N}.

Put t=1βˆ’xt = 1 - x, so x=1βˆ’tx = 1 - t and dx=βˆ’dtdx = -dt. The limits swap (x=0β‡’t=1x=0 \Rightarrow t=1, x=1β‡’t=0x=1 \Rightarrow t=0):

∫01x(1βˆ’x)n dx=∫01(1βˆ’t) tn dt=∫01(tnβˆ’tn+1)dt.\int_0^1 x(1-x)^n\,dx = \int_0^1 (1-t)\,t^n\,dt = \int_0^1 \left(t^n - t^{n+1}\right) dt.

Integrate:

=[tn+1n+1βˆ’tn+2n+2]01=1n+1βˆ’1n+2=(n+2)βˆ’(n+1)(n+1)(n+2)=1(n+1)(n+2).= \left[\frac{t^{n+1}}{n+1} - \frac{t^{n+2}}{n+2}\right]_0^1 = \frac{1}{n+1} - \frac{1}{n+2} = \frac{(n+2)-(n+1)}{(n+1)(n+2)} = \frac{1}{(n+1)(n+2)}.

(Check n=1n=1: ∫01x(1βˆ’x) dx=12βˆ’13=16=12β‹…3\int_0^1 x(1-x)\,dx = \tfrac12 - \tfrac13 = \tfrac16 = \tfrac{1}{2\cdot 3}.)

βœ“Final answer

∫01x(1βˆ’x)n dx=1(n+1)(n+2).\displaystyle \int_0^1 x(1-x)^n\,dx = \frac{1}{(n+1)(n+2)}.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.