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If I_{n}=\int_{0}^{1}e^{x}\left ( x-1 \right )^{n}dx, then which of the following statement is correct?

Option: 1

(1)\; I_{n} = (-1)^{n-1}-nI_{n-1}


Option: 2

(2)\; I_{n} = 1-nI_{n-1}


Option: 3

(3)\; I_{n} = \left ( -1 \right )^{n+1}-\left ( n+1 \right )I_{n-1}

 


Option: 4

(4)\; I_{n} = \left ( -1 \right )^{n}-nI_{n-1}


Answers (1)

best_answer

I_{n}=\int_{0}^{1}e^{x}\left ( x-1 \right )^{n}dx

                    II         I

I_{n}=\left [ \left ( x-1 \right )^{n}e^{x} \right ]_{0}^{1}-\int_{0}^{1}n\left ( x-1 \right )^{n-1}.e^{x}dx

I_{n}=-\left ( -1 \right )^{n}-n\; I_{n-1}=\left ( -1 \right )^{n-1}-nI_{n-1}

Posted by

vishal kumar

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