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  • Need clarity, kindly explain! - Integral Calculus - JEE Main-11

Evaluate  \int_{0}^{\infty} e^{-x^{3}}\;dx.

  • Option 1)

    \sqrt{\frac{1}{3}}

  • Option 2)

    \sqrt{\frac{2}{3}}

  • Option 3)

    \sqrt{\frac{4}{3}}

  • Option 4)

    \sqrt{\pi}

 

Answers (1)

best_answer

As we have learnt,

 

Gamma Function -

If n is positive rational number then

\int_{o}^{\infty }e^{-x}x^{n-1}dx

 

- wherein

Γ (n) is known as Gamma Function

 

 

I = \int_{0}^{infty} e^{-x^{3}}\;dx

Put

 x^{3} = t \Rightarrow 3x^{2}dx = dt \Rightarrow dx = \frac{dt}{3x^2} = t^{-\frac{2}{3}}\frac{dt}{3}

Thus,

I = \int_{0}^{\infty}\frac{1}{3}t^{-\frac{2}{3}}e^{-t}dt = \frac{1}{3}\Gamma(\frac{1}{3}) = \left (\frac{4}{3} - 1\right) \Gamma \left (\frac{4}{3} -1\right ) = \Gamma(\frac{4}{3})


Option 1)

\sqrt{\frac{1}{3}}

Option 2)

\sqrt{\frac{2}{3}}

Option 3)

\sqrt{\frac{4}{3}}

Option 4)

\sqrt{\pi}

Posted by

gaurav

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