Question

Let $I_n={\int }_0^1(1-x^3{)}^{n}dx,(n\in N)$ then

• A. $3n{I}_n=(3n-1)I_{n-1}\forall n\ge 2$
• B. $(3n-1)I_n=3n{I}_{n-1}\forall n\ge 2$
• C. $(3n-1)I_n=(3n+1)I_{n-1}\forall n\ge 2$
• D. $(3n+1)I_n=3n{I}_n\forall n\ge 2$

Answer 1

The correct option is D $(3n+1)I_n=3n{I}_n\forall n\ge 2$

Consider the given function,

let

$I_n={\int }_0^1(1-x^3{)}^{n}dx,n\in N$

$I_n={\int }_0^1\begin{array}{cc}1.& {(1-x^3)}^{n}dx,n\in N\\ IInd& Ist\end{array}$

$I_n=(1-x^3{)}^n{\int }_0^{1}1dx-{\int }_0^1\left[\frac{d}{dx}\right(1-x^3{)}^n{\int }_0^{1}1dx]dx$

$I_n={\left[x\right(1-x^3{)}^n]}_0^1+3n{\int }_0^1{x}^3(1-x^3{)}^{n-1}dx$

$=0+3n{\int }_0^1[1-(1-x^3\left)\right)(1-x^3{)}^{n-1}]dx$

$I_n=3n{I}_{n-1}-3n{I}_n$

$I_n+3n{I}_n=3n{I}_{n-1}$

$I_n(1+3n)=3n{I}_n,n\in N$

$(3n+1)I_n=3n{I}_n,\forall n\ge 2$

Hence, this is the answer.