Integration Challange

Let \[\displaystyle{{ I }_{ 1 }=\int _{ 0 }^{ x }{ { e }^{ zx }{ e }^{ -{ z }^{ 2 } } } dz\\ { I }_{ 2 }=\int _{ 0 }^{ x }{ { e }^{ -\cfrac { { z }^{ 2 } }{ 4 } }dz } \\ \boxed { \cfrac { { I }_{ 1 } }{ { I }_{ 2 } } =? } }\]

#Calculus #Integration #JEE

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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`\boxed{123}`	$\boxed{123}$

Comments

@Brian Charlesworth @Kunal Joshi @Ronak Agarwal @Raghav Vaidyanathan @Calvin Lin

Nishu sharma - 6 years, 1 month ago

@Nishu sharma

The answer is:

$\displaystyle \large \large e^{\frac{x^2}{4}}$

Raghav Vaidyanathan - 6 years, 1 month ago

Yes ! But I gave this as a Challange , So Everyone should post thier approaches .. Like karthik did !

Nishu sharma - 6 years, 1 month ago

Well, it's very easy. And Raghav has given exactly the correct answer considering it to be very easy.

$\displaystyle {I}_{1} = \int_{0}^{x}{{e}^{-{z}^{2}+zx}}$

Completing the square,

$\displaystyle {I}_{1} = \int_{0}^{x}{{e}^{{x}^{2}/4}{e}^{-{z}^{2}+zx+{x}^{2}/4}}$

$\displaystyle {I}_{1} = {e}^{{x}^{2}/4}\int_{0}^{x}{{e}^{-{(z-x/2)}^{2}}}$

Using u-substitution and a little bashing up, we get

$\displaystyle {I}_{1} = {e}^{{x}^{2}/4}\int_{-x/2}^{x/2}{{e}^{-{u}^{2}}}$

And hence, $\displaystyle {I}_{1} = 2{e}^{{x}^{2}/4}\int_{0}^{x/2}{{e}^{-{u}^{2}}}$

By using error function now,

$\displaystyle {I}_{1} = {e}^{{x}^{2}/4}\sqrt{\pi}\text{erf}(x/2)$

Now, $\displaystyle {I}_{2} = \int_{0}^{x}{{e}^{-{(z/2)}^{2}}}$

Using v-substitution and some 'using of facts',

$\displaystyle {I}_{2} = 2\int_{0}^{x/2}{{e}^{-{v}^{2}}}$

$\displaystyle {I}_{2} = \sqrt{\pi}\text{erf}(x/2)$

Therefore, $\displaystyle \frac{{I}_{1}}{{I}_{2}} = {e}^{{x}^{2}/4}$

Kartik Sharma - 6 years, 1 month ago

Kartik there was no need to uneccasarily bring the error function into play here.

Ronak Agarwal - 6 years, 1 month ago

Yes, you can convert one integral directly into the other without erf().

Raghav Vaidyanathan - 6 years, 1 month ago

Yeah, I agree. That was a mistake!

Kartik Sharma - 6 years, 1 month ago

Hi Ronak ... I have solved Your Physics questions , I found all of them very Interesting . I really much impressed from your Problem making skills . I really appreciate awesomeness of your Problems ! Great work .. ! I'am regret that I joined brilliant very late , My relative brother recommended this site to me recently ... There are so many good questions of maths and physics , and I found that :: You and Jatin yadav and Deepanshu and David Mattingly and Josh silverman Posted so many self made questions .. I'am really amazed with your skills !

Nishu sharma - 6 years, 1 month ago

@Nishu Sharma – I know it's pretty late to type this but thanks for the appreciation, I see you are from jaipur also, in which area of jaipur are you from, I am from jaipur too. @Nishu sharma

Ronak Agarwal - 6 years, 1 month ago

@Ronak Agarwal – Hi , Actually My home is at jaipur , but I live in kota alone , near by to my aunt , My family lives in Jaipur. My home in backyard of Vidhan-sabha .. where do you live ?

Nishu sharma - 6 years, 1 month ago

Also, never miss out $dz$ inside integral. It is a bad habit.

Raghav Vaidyanathan - 6 years, 1 month ago

Yeah. But it is tedious. :P

Kartik Sharma - 6 years, 1 month ago

Nicely done , but there is more Smarter approach for this .. Think about it Kartik sharma

Nishu sharma - 6 years, 1 month ago

Can anyone explain this https://brilliant.org/discussions/thread/extremely-weird-integration/?ref_id=946409

柯南 - 5 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$