If In =∫xnecx dx for n≥1 then cIn+nIn−1 is equal to

In=∫xnecxdxIn=ecxcxn−∫ecxc.nxn−1dxIn=ecxxnc−ncIn−1cIn+nIn−1=ecx.xn

If In =∫xnecx dx for n≥1 then cIn+nIn−1 is equal to

In=∫xnecxdxIn=ecxcxn−∫ecxc.nxn−1dxIn=ecxxnc−ncIn−1cIn+nIn−1=ecx.xn

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If $I_{n} = \int x^{n} e^{c x} d x f o r n \geq 1 t h e n c I_{n} + n I_{n - 1} i s e q u a l t o$

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$x^{n} + e^{c x}$

$x^{n} e^{c x}$

$e^{c x}$

$x^{n}$

answer is A.

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Detailed Solution

$I_{n} = \int x^{n} e^{c x} d x$
$I_{n} = \frac{e^{c x}}{c} x^{n} - \int \frac{e^{c x}}{c} . n x^{n - 1} d x$
$I_{n} = \frac{e^{c x} x^{n}}{c} - \frac{n}{c} I_{n - 1}$
$c I_{n} + n I_{n - 1} = e^{c x} . x^{n}$

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If In =∫xnecx dx for n≥1 then cIn+nIn−1 is equal to

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If $I_{n} = \int x^{n} e^{c x} d x f o r n \geq 1 t h e n c I_{n} + n I_{n - 1} i s e q u a l t o$