Integration of e^x²
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Answered by
2
we can't integrate directly ∫eˣ².dx
use taylor series ,
expansion of eˣ² = 1 ₊ x² ₊ x⁴/2 + x⁶/6 + x⁸/24 +..............∞
now use this in place of eˣ² and then integrate by using fundamental concept of integration.
∫eˣ².dx = ∫( 1 + x² + x⁴/2 + x⁶/6 + x⁸/24 +............∞)dx
=∫1.dx +∫x².dx +1/2∫x⁴.dx..............∞)dx
=x + x³/3 +x⁵/10 +.................∞
use taylor series ,
expansion of eˣ² = 1 ₊ x² ₊ x⁴/2 + x⁶/6 + x⁸/24 +..............∞
now use this in place of eˣ² and then integrate by using fundamental concept of integration.
∫eˣ².dx = ∫( 1 + x² + x⁴/2 + x⁶/6 + x⁸/24 +............∞)dx
=∫1.dx +∫x².dx +1/2∫x⁴.dx..............∞)dx
=x + x³/3 +x⁵/10 +.................∞
Anonymous:
and who verified it ?
@cosmick read answer sincerely . i said its not integrate directly , but we integrate by using expansion of given function
:( don't scold me
tell me why integration of √sinx is not possible
i didn't scold you bro , i just say you.
don't do spell mistake in my name too
who says √sinx is not integrable
come in dm
is it ?
Answered by
3
Hi Friend,
Here is your answer,
eˣ².dx
We will use taylor series,
We will do expansion of eˣ² => 1 ₊ x² ₊ x⁴/2 + x⁶/6 + x⁸/24 +................. infinite
)
eˣ².dx = ( 1 + x² + x⁴/2 + x⁶/6 + x⁸/24 +......infinite) dx
=>1.dx +x².dx +1/2 x⁴.dx..........infinte) dx
=>x + x³/3 +x⁵/10 +............infinite)
Hope it helps you!
Here is your answer,
eˣ².dx
We will use taylor series,
We will do expansion of eˣ² => 1 ₊ x² ₊ x⁴/2 + x⁶/6 + x⁸/24 +................. infinite
)
eˣ².dx = ( 1 + x² + x⁴/2 + x⁶/6 + x⁸/24 +......infinite) dx
=>1.dx +x².dx +1/2 x⁴.dx..........infinte) dx
=>x + x³/3 +x⁵/10 +............infinite)
Hope it helps you!
sorry bhai but its wrong
it is not integrable
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