New user? Sign up

Existing user? Log in

Find the value of the definite integral : \[ \displaystyle \int_0^1( 1+e^{-x^{2}} )dx \]

Note by Rishu Jaar 7 months, 2 weeks ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

> This is a quote

This is a quote

# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

Sort by:

Read Error function.

Log in to reply

@Pi Han Goh , thank you , its a jee 1981 question but didn't knew the maths involved was so higher.

Then you can only evaluate it using numerical methods, like trapezoidal rule or Simpson's rule.

@Pi Han Goh – Thanks , could you provide a link?

@Rishu Jaar – Trapezium Rule

@Pi Han Goh – Thanks!

If, by value, you mean the area under the curve from 0 to 1, the solution would be 1/2 sqrt(π) • e • rf(1) + 1. This is simply found by removing the parentheses, taking the derivative, and then solving from there as usual.

Oh can you explain more.

Yes - I'll post a new discussion called JEE 1981 Int Calc with the information enclosed. I need my LaTeX!

@Hunter Edwards – Ok sure.

@Rishu Jaar – @RISHU Jaar It's in the new section.

@Hunter Edwards – Done.

@Rishabh Cool and others please give a proof !

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestRead Error function.

Log in to reply

@Pi Han Goh , thank you , its a jee 1981 question but didn't knew the maths involved was so higher.

Log in to reply

Then you can only evaluate it using numerical methods, like trapezoidal rule or Simpson's rule.

Log in to reply

Log in to reply

Trapezium Rule

Log in to reply

Log in to reply

If, by value, you mean the area under the curve from 0 to 1, the solution would be 1/2 sqrt(π) • e • rf(1) + 1. This is simply found by removing the parentheses, taking the derivative, and then solving from there as usual.

Log in to reply

Oh can you explain more.

Log in to reply

Yes - I'll post a new discussion called JEE 1981 Int Calc with the information enclosed. I need my LaTeX!

Log in to reply

Log in to reply

@RISHU Jaar It's in the new section.

Log in to reply

Log in to reply

@Rishabh Cool and others please give a proof !

Log in to reply