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A relatively simple integral problem...

Bmanmcfly

Member
Mar 10, 2013
42
Hi again,
so, I've been blasting away at the integration problems that I've been facing and just when I thought this term was going too easy, then I got to this one that kicked me in the left tooth.

Problem: \(\displaystyle \int_{0}^{1}4\ln(e^x)e^{-2x^2}dx\)

Attempted solution:
Ok, first off we move the 4 to the other side of the integral.

Now, I find \(\displaystyle u=e^{-2x^2}\frac{du}{dx}=\frac{-4x*\ln(e^{-2x^2})}{e^x}dx\)

\(\displaystyle =\frac{4}{-4}\int_{0}^{1}\frac{\ln(e^{-2x^2})}{e^x}\)
\(\displaystyle =-(\frac{\ln(e^{-2(1)^2})}{e^{(1)}}-\frac{\ln(e^{-2(0)^2})}{e^{(0)}})\)

Is this correct so far? It feels wrong, but I can't seem to find anywhere that specifically shows the integral of lne^x.

Thankd for any help...


 
Last edited by a moderator:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
We are given to evaluate:

\(\displaystyle I=\int_0^1 4\ln\left(e^x \right)e^{-2x^2}\,dx\)

This first thing I notice is \(\displaystyle \ln\left(e^x \right)=x\) so I would rewrite:

\(\displaystyle I=\int_0^1 4xe^{-2x^2}\,dx\)

Next I would let:

\(\displaystyle u=-2x^2\,\therefore\,du=-4x\) and we have:

\(\displaystyle I=-\int_0^{-2} e^{u}\,du=\int_{-2}^{0} e^{u}\,du\)
 

Bmanmcfly

Member
Mar 10, 2013
42
We are given to evaluate:

\(\displaystyle I=\int_0^1 4\ln\left(e^x \right)e^{-2x^2}\,dx\)

This first thing I notice is \(\displaystyle \ln\left(e^x \right)=x\) so I would rewrite:
Really?? So the \(\displaystyle \ln\left(e^x \right)\) was basically a trick to see how much you were paying attention....
\(\displaystyle I=\int_0^1 4xe^{-2x^2}\,dx\)

Next I would let:

\(\displaystyle u=-2x^2\,\therefore\,du=-4x\) and we have:

\(\displaystyle I=-\int_0^{-2} e^{u}\,du=\int_{-2}^{0} e^{u}\,du\)
I'm curious, why is the int going from 0 to -2?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
When I rewrote the integral in terms of the new variable, the limits have to be changed in accordance with the substitution:

\(\displaystyle u(x)=-2x^2\)

hence:

\(\displaystyle u(0)=0\)

\(\displaystyle u(1)=-2\)
 

Bmanmcfly

Member
Mar 10, 2013
42
When I rewrote the integral in terms of the new variable, the limits have to be changed in accordance with the substitution:

\(\displaystyle u(x)=-2x^2\)

hence:

\(\displaystyle u(0)=0\)

\(\displaystyle u(1)=-2\)
Ok, I think I get it... But is this a matter of simplification as it seems, or necessary?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
For definite integrals, when I make a substitution, I like to leave behind the old variable completely, so I don't have to back-substitute at the end. I just find it simpler and more elegant.
 

Bmanmcfly

Member
Mar 10, 2013
42
That's actually a good technique I'll start doing that.

Thanks again for the help, I thought I would find a similar simple example, but no luck.

And the following questions carry on as a breeze... Which worries me, cause of its too easy I feel like in doing something wrong...
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
And the following questions carry on as a breeze... Which worries me, cause of its too easy I feel like in doing something wrong...
Sometimes calculus problems are very easy. And sometimes they are very difficult. That's the nature of a mathematical structure as immensely powerful as calculus.

So figure out ways to check your answer. Checking your answer always involves using an independent method of solving the problem, or perhaps estimating the solution, or working backwards from the solution to the problem statement. Your philosophy should be this: the answer is not correct until it is checked. (Repeat that like a mantra to yourself.) Your checking method has to be independent from the original way of solving the problem, because you've already thought that through, and it's correct, right? Uh, right?

For example, to check an indefinite integral, differentiate your answer and see if you get the original integrand back to you. For a definite integral, treat it like the area problem that it is, and use straight lines to approximate the solution (perhaps get an upper and a lower bound on the actual area).