Difficult 2u integration of exponentials - HELP please (1 Viewer)

[Lucas_]

Hey,

I have been asked: <a href="http://www.codecogs.com/eqnedit.php?latex=Find \int x^2 e^{x}^{3-1} dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?Find \int x^2 e^{x}^{3-1} dx" title="Find \int x^2 e^{x}^{3-1} dx" /></a>

That x is raised again to (3-1). So x^2 e^x^3-1. I couldn't figure out how to get latex to raise it to another power properly.

I have no idea how to do this. Somebody please help

 

[Peeik]

Did you make a typo in the question? Are there meant to be brackets around the exponent on the exponential??

 

[Shadowdude]

x3x-1?

 

[Lucas_]

That x is raised again to (3-1). So x^2 e^x^3-1. I couldn't figure out how to get latex to raise it to another power properly.

 

[thorax94]

That x is raised again to (3-1). So x^2 e^x^3-1. I couldn't figure out how to get latex to raise it to another power properly.

Edit: read the question wrong:

Let u = x^3-1
Du= 3x^2dx

Therefore integration becomes :
1/3 times integration of e^u
= (e^u)/3
= e^(x^3-1)/3 + c

 

[Shadowdude]

?

 

[Timske]

<a href="http://www.codecogs.com/eqnedit.php?latex=\int x^2e^{x^{3-1}} dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int x^2e^{x^{3-1}} dx" title="\int x^2e^{x^{3-1}} dx" /></a>

?

 

[Lucas_]

?

Yep this is it. Sorry, textbook is very unclear with its small writing.

 

[deswa1]

I think the question is:

<a href="http://www.codecogs.com/eqnedit.php?latex=\int x^2e^{x^3-1}dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int x^2e^{x^3-1}dx" title="\int x^2e^{x^3-1}dx" /></a>

Hint: Differentiate <a href="http://www.codecogs.com/eqnedit.php?latex=e^{x^3-1}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?e^{x^3-1}" title="e^{x^3-1}" /></a>

and see what you get

 

[J-Wang]

hey lucas. i think i have done that question. the X on the e is raised to the power of 3 and the -1 is just hanging on the edge. ie e^x^3-1 ie x cubed -1. should be able 2 do it from there

 

[J-Wang]

use a u substitution

 

[RealiseNothing]

Use the fact that a^b^c = a^bc.

Hence the question becomes:



Which is:



You should be able to do it from here.

EDIT: OP changed question :/

 

[Sanjeet]

I think the question is:

<a href="http://www.codecogs.com/eqnedit.php?latex=\int x^2e^{x^3-1}dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int x^2e^{x^3-1}dx" title="\int x^2e^{x^3-1}dx" /></a>

Hint: Differentiate <a href="http://www.codecogs.com/eqnedit.php?latex=e^{x^3-1}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?e^{x^3-1}" title="e^{x^3-1}" /></a>

and see what you get

Exactly how you do it.

 

[barbernator]

use a u substitution

2 unit man, they don't do subs. do what deswa said

 

[thorax94]

Made it pretty for you:
<a href="http://www.codecogs.com/eqnedit.php?latex=\LARGE \int%20x^{2}e^{x^{3}-1} \; Let \: u \: = \: x^3-1\Rightarrow du=3x^2dx\Rightarrow equation \: becomes \:\frac{1}{3} \int e^u\Rightarrow \frac{e^u}{3}\Rightarrow \frac{e^{3x-1}}{3}@plus;c" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\LARGE \int%20x^{2}e^{x^{3}-1} \; Let \: u \: = \: x^3-1\Rightarrow du=3x^2dx\Rightarrow equation \: becomes \:\frac{1}{3} \int e^u\Rightarrow \frac{e^u}{3}\Rightarrow \frac{e^{3x-1}}{3}+c" title="\LARGE \int%20x^{2}e^{x^{3}-1} \; Let \: u \: = \: x^3-1\Rightarrow du=3x^2dx\Rightarrow equation \: becomes \:\frac{1}{3} \int e^u\Rightarrow \frac{e^u}{3}\Rightarrow \frac{e^{3x-1}}{3}+c" /></a>

 

[Shadowdude]

Made it pretty for you:
<a href="http://www.codecogs.com/eqnedit.php?latex=\LARGE \int%20x^{2}e^{x^{3}-1} \; Let \: u \: = \: x^3-1\Rightarrow du=3x^2dx\Rightarrow equation \: becomes \:\frac{1}{3} \int e^u\Rightarrow \frac{e^u}{3}\Rightarrow \frac{e^{3x-1}}{3}@plus;c" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\LARGE \int%20x^{2}e^{x^{3}-1} \; Let \: u \: = \: x^3-1\Rightarrow du=3x^2dx\Rightarrow equation \: becomes \:\frac{1}{3} \int e^u\Rightarrow \frac{e^u}{3}\Rightarrow \frac{e^{3x-1}}{3}+c" title="\LARGE \int%20x^{2}e^{x^{3}-1} \; Let \: u \: = \: x^3-1\Rightarrow du=3x^2dx\Rightarrow equation \: becomes \:\frac{1}{3} \int e^u\Rightarrow \frac{e^u}{3}\Rightarrow \frac{e^{3x-1}}{3}+c" /></a>

2u doesn't do substitution in integration

 

[thorax94]

2u doesn't do substitution in integration

My apologies. I have no idea how to approach the question then.

p.s how do you start a new line in LATEX?

 

[Sanjeet]

Which is:



You should be able to do it from here.

EDIT: OP changed question :/

How exactly can you do that with 2u knowledge?

 

[cutemouse]

How exactly can you do it from there with 2u knowledge?

d/dx e^f(x) = f'(x) e^f(x)

So integral f'(x) e^f(x) dx = e^f(x) + c

So you recognise the "pattern" in 2U.

 

[thorax94]

Any chance this is a trapezoidal/Simpson's rule question?