If ∫$$0$$∞ $$x2n+1$$⋅e−$$x2dx=360$$ then the value of n is

$$I=$$∫$$0$$∞ $$x2n$$⋅xe−$$x2dxPut$$ $$x2=t$$ and $$xdx=dt/2$$ ∴ $$I=12$$∫$$0$$∞ tne−$$tdt=12$$−tne−$$t0$$∞$$+n$$∫$$0$$∞ tn−$$1e$$−$$tdt=120+n$$∫$$0$$∞ tn−$$1e$$−$$tdt=n$$!$$2=360$$⇒ $$n=6$$

Evaluation of definite integrals

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Question

If $\int_{0}^{\infty} x^{2 n + 1} \cdot e^{- x^{2}} d x = 360$ then the value of $n$ is

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Solution

$I = \int_{0}^{\infty} {(x^{2})}^{n} \cdot x e^{- x^{2}} d x$
Put $x^{2} = t$ and $x d x = d t / 2$
$\begin{array}{l} ∴ I = \frac{1}{2} \int_{0}^{\infty} t^{n} e^{- t} d t \\ = \frac{1}{2} {[- t^{n} e^{- t}]}_{0}^{\infty} + n \int_{0}^{\infty} t^{n - 1} e^{- t} d t] \\ = \frac{1}{2} [0 + n \int_{0}^{\infty} t^{n - 1} e^{- t} d t] \\ = \frac{n!}{2} = 360 \\ \Rightarrow n = 6 \end{array}$

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Evaluation of definite integrals

Remember concepts with our Masterclasses.

If ∫0∞ x2n+1⋅e−x2dx=360 then the value of n is

I=∫0∞ x2n⋅xe−x2dxPut x2=t and xdx=dt/2 ∴ I=12∫0∞ tne−tdt=12−tne−t0∞+n∫0∞ tn−1e−tdt=120+n∫0∞ tn−1e−tdt=n!2=360⇒ n=6

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If $\int_{0}^{\infty} x^{2 n + 1} \cdot e^{- x^{2}} d x = 360$ then the value of $n$ is