Problem: Definite Integral with (2x^2015)(e^-|x|)

This will get pretty ugly, but I think the best way to proceed after separating the integral is to use integration by parts 2015 times.

\(\displaystyle \int _{-1}^1\:\left(1+2x^{2015}\right)\cdot \:e^{-\left|x\right|}dx=\int _{-1}^0\:\left(1+2x^{2015}\right)\cdot e^xdx+\int _0^1\:\left(1+2x^{2015}\right)\cdot e^{-x}dx\)

Take the first integral and start...

\(\displaystyle u=1+2x^{2015}\); \(\displaystyle du=4030x^{2014}dx\)
\(\displaystyle dv=e^{x}\); \(\displaystyle v=e^{x}\)

\(\displaystyle \left(-\frac{1}{e}-1\right)-\int _{-1}^0\:4030x^{2014}\cdot e^xdx\)

\(\displaystyle u=4030x^{2014}\); \(\displaystyle du=(4030\cdot2014)x^{2013}dx\)
\(\displaystyle dv=e^{x}\); \(\displaystyle v=e^{x}\)

And so on. Eventually, you'll notice a pattern in the repeated derivatives which allows you to skip some steps, but it's still not going to be pretty. And then, of course, you'll need to deal with the upper integral in the same manner.
 
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Vect said:
https://www.wolframalpha.com/input/?i=integral+from+-1+to+1+(1+2x^2015)e^-|x|
https://www.desmos.com/calculator/udrwtflrbw

First step is to express the module of x and separate the integral in two parts. Then induction? Obviously, the answer is between 1 and 2.

A pointer in the right direction from someone experienced with calculus is all I need.

Thanks in advance!
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Split the integral as ksdhart did. Put out the 1 part of the 1+2x2015 since those are just an integral of \(\displaystyle e^{\pm\, x}\). Now look at
2 \(\displaystyle \int_{-1}^0\, x^{2015}\, e^{-x}\, dx\)
What would happen if you did a u=-x substitution? What does that tell you about any odd exponent? What happen with even exponents?
 
Ishuda said:
Split the integral as ksdhart did. Put out the 1 part of the 1+2x2015 since those are just an integral of \(\displaystyle e^{\pm\, x}\). Now look at 2 \(\displaystyle \int_{-1}^0\, x^{2015}\, e^{-x}\, dx\)
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What are you saying?
The statement "integral of \(\displaystyle \Large\color{red}{e^{\pm\, x}}\)" makes no sense. Does it?

Are you saying \(\displaystyle \displaystyle\int_{-1}^0\, x^{2015}\, e^{x}\, dx+\int_{0}^1\, x^{2015}\, e^{-x}\, dx\) can be rewritten as a single integral?

Is the original question about an odd function?
 
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pka said:
What are you saying?
The statement "integral of \(\displaystyle \Large\color{red}{e^{\pm\, x}}\)" makes no sense. Does it?

Are you saying \(\displaystyle \displaystyle\int_{-1}^0\, x^{2015}\, e^{x}\, dx+\int_{0}^1\, x^{2015}\, e^{-x}\, dx\) can be rewritten as a single integral?

Is the original question about an odd function?
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The "integral from -1 to 1 (1+2x^2015)e^-|x|" can be broken into four parts:
(1)\(\displaystyle \int_{-1}^0\, e^x\, dx\)
(2)\(\displaystyle \int_0^1\, e^{-x}\, dx\)
(3)2 \(\displaystyle \int_{-1}^0\, x^{2015}\, e^x\, dx\)
(4)2 \(\displaystyle \int_0^1\, x^{2015}\, e^{-x}\, dx\)

I got lazy when I said "those are just an integral of \(\displaystyle e^{\pm\, x}\)" What I meant by that was that integrals (1) and (2) are just the integral of either ex or e-x which are simple integrals and easy to do.

Take my hint and let u=-x in (3). then dx =-du, x=-1 --> u=1, x=0 --> u=0. Thus
\(\displaystyle \int_{-1}^0\, x^{2015}\, e^x\, dx\) = \(\displaystyle \int_{1}^0\, (-u)^{2015}\, e^{-u}\, (-du)\, =\, \int_{1}^0\, u^{2015}\, e^{-u}\, du\, =\, -\int_0^1\, x^{2015}\, e^{-x}\, dx\)
So, yes, you might say that Equations (3) and (4) can be combined and written as a single integral. Then again, you might just note that they are the negative of one another and their sum adds to zero and what you are left with is (1) and (2).

As far as the questions about the odd and even numbers: It was just a statement to get one to notice that for any odd exponent, the same substitution works whereas for an even number you just end up with basically the same problem again [but without the absolute value sign and with an interval change].
 
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Ishuda said:
The "integral from -1 to 1 (1+2x^2015)e^-|x|" can be broken into four parts:
(1)\(\displaystyle \int_{-1}^0\, e^x\, dx\)
(2)\(\displaystyle \int_0^1\, e^{-x}\, dx\)
(3)2 \(\displaystyle \int_{-1}^0\, x^{2015}\, e^x\, dx\)
(4)2 \(\displaystyle \int_0^1\, x^{2015}\, e^{-x}\, dx\).
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I hate to ruin a good question that some competent mathematics teacher carefully designed. BUT
All that is above is wasted effort. Reading the original posting, it is clear that the question is about odd & even functions.

\(\displaystyle \Large\displaystyle{\int_{ - 1}^1 {\left( {1 + 2{x^{2015}}} \right){e^{ - \left| x \right|}}dx} = 2\int_0^1 {{e^{ - x}}dx}}\)
 
pka said:
I hate to ruin a good question that some competent mathematics teacher carefully designed. BUT
All that is above is wasted effort. Reading the original posting, it is clear that the question is about odd & even functions.

\(\displaystyle \Large\displaystyle{\int_{ - 1}^1 {\left( {1 + 2{x^{2015}}} \right){e^{ - \left| x \right|}}dx} = 2\int_0^1 {{e^{ - x}}dx}}\)
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I don't think the given integrand (as a whole) is an even function (neither it is odd).

For x < 1, it behaves almost like an even function - because |x2015|<< 1

This will be apparent, when we try to plot the function beyond x = 1

I think, you and Ishuda are stressing the same point:

Part of the integrand is an even function and part of it is odd function. Ishuda (I believe) was trying to cajole the OP to come to same conclusion.
 
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Subhotosh Khan said:
I don't think the given integrand (as a whole) is an even function (neither it is odd).

For x < 1, it behaves almost like an even function - because |x2015|<< 1

This will be apparent, when we try to plot the function beyond x = 1

I think, you and Ishuda are stressing the same point:

Part of the integrand is an even function and part of it is odd function. Ishuda (I believe) was trying to cajole the OP to come to same conclusion.
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Thank You.
 
Subhotosh Khan said:
I don't think the given integrand (as a whole) is an even function (neither it is odd).
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Sorry but that is not the case.
Break up the sum: \(\displaystyle \Large\displaystyle{\int\limits_{ - 1}^1 {\left( {1 + 2{x^{2015}}} \right){e^{ - \left| x \right|}}dx} = \int\limits_{ - 1}^1 {{e^{ - \left| x \right|}}dx} + \int\limits_{ - 1}^1 {2{x^{2015}}{e^{ - \left| x \right|}}dx}}\)

Now consider the two intergrands, \(\displaystyle {\Large E(x) = {e^{ - \left| x \right|}}\quad \& \quad O(x) = 2{x^{2015}}{e^{ - \left| x \right|}}}\).

Is it not clear that \(\displaystyle {\Large E(x)=E(-x)~\&~ O(-x) = -O(x)}~?\). One EVEN and one ODD

\(\displaystyle \Large\displaystyle\int\limits_{ - 1}^1 {{e^{ - |x| }}dx} = 2\int\limits_0^1 {{e^{ - x}}dx}~~\&~~\Large\displaystyle\int\limits_{ - 1}^1 {2{x^{2015}}{e^{ - \left| x \right|}}dx}=0 \)

This problem is designed to force people who do not see the trick to waste time and not finish the test set. Test-Prep folks teach students to lookout for that all the time. This question is just too complicated to not fit in the "trick" category.
 
pka said:
Subhotosh Khan said:
I don't think the given integrand (as a whole) is an even function (neither it is odd).
Click to expand...
Sorry but that is not the case.
Break up the sum: \(\displaystyle \Large\displaystyle{\int\limits_{ - 1}^1 {\left( {1 + 2{x^{2015}}} \right){e^{ - \left| x \right|}}dx} = \int\limits_{ - 1}^1 {{e^{ - \left| x \right|}}dx} + \int\limits_{ - 1}^1 {2{x^{2015}}{e^{ - \left| x \right|}}dx}}\)

Now consider the two intergrands, \(\displaystyle {\Large E(x) = {e^{ - \left| x \right|}}\quad \& \quad O(x) = 2{x^{2015}}{e^{ - \left| x \right|}}}\).

Is it not clear that \(\displaystyle {\Large E(x)=E(-x)~\&~ O(-x) = -O(x)}~?\). One EVEN and one ODD

\(\displaystyle \Large\displaystyle\int\limits_{ - 1}^1 {{e^{ - |x| }}dx} = 2\int\limits_0^1 {{e^{ - x}}dx}~~\&~~\Large\displaystyle\int\limits_{ - 1}^1 {2{x^{2015}}{e^{ - \left| x \right|}}dx}=0 \)

This problem is designed to force people who do not see the trick to waste time and not finish the test set. Test-Prep folks teach students to lookout for that all the time. This question is just too complicated to not fit in the "trick" category.
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We all, you, Subhotosh Khan, and I (and others, I'm sure), agree that E(x)=e-|x| is even and O(x)=x2015 e-|x| is odd. So, with your statement of "Sorry but that is not the case." when you quoted just the one part of Subhotosh Khan's post "I don't think the given integrand (as a whole) is an even function (neither it is odd).", what is the integrand function
F(x) = E(x) + 2 O(x)
Is it even or odd? Or is there another choice besides even/not even, odd/not odd, neither even nor odd?

If you consider 0 an even number (0 = 2 * 0), then it seems that, if one though about it (as apparently some people don't), then
one would realize from my post the e-|x| = x0 e-|x| was an even function and x2015 e-|x| was an odd function. If fact, given that n is an integer, En(x) = x2n e-|x| is an even function and On(x) = x2n+1 e-|x| is an odd function. Also that
\(\displaystyle \int_{-x}^{x}E_n(x)\, dx\, = 2\, \int_0^x\, E_n(x)\, dx\)
and
\(\displaystyle \int_{-x}^{x}O_n(x)\, dx\, = -\int_0^x\, O_n(x)\, dx\, +\, \int_0^x\, O_n(x)\, dx\, =\, 0 \)
 
Ishuda said:
Is it even or odd? Or is there another choice besides even/not even, odd/not odd, neither even nor odd?

If you consider 0 an even number (0 = 2 * 0), then it seems that, if one though about it (as apparently some people don't),
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With that, I know you are trolling. Or else you need to study some mathematics..

Can you prove that any function is the sum of an even function and an odd function?
 
pka said:
With that, I know you are trolling. Or else you need to study some mathematics..

Can you prove that any function is the sum of an even function and an odd function?
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Well, as usual, you have gotten down to changing the subject when you can't answer the questions asked. This time though, I do have a final answer which is really another question for you: Why do like to challenge someone by making comments which seem to show people incompetent rather than hold a civilized conversation by just pointing out mistakes and possibly even point it out privately?

I'm really not expecting an answer to the question and even if you do, I am through with this thread since nothing of interest to the OP's question seems to be happening any more.
 
Ishuda said:
Why do like to challenge someone by making comments which seem to show people incompetent rather than hold a civilized conversation by just pointing out mistakes and possibly even point it out privately?
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That is easy to answer: Amateur arguments drive me nuts.

\(\displaystyle \Large{f(x)=\dfrac{f(x)+f(-x)}{2}+\dfrac{f(x)-f(-x)}{2}}\)
 
pka said:
That is easy to answer: Amateur arguments drive me nuts.

\(\displaystyle \Large{f(x)=\dfrac{f(x)+f(-x)}{2}+\dfrac{f(x)-f(-x)}{2}}\)
Click to expand...

Just couldn't resist. It's not a drive, just an extremely short walk. BTW: That's what I'm often told. So, although you may not like the company, you are not alone.
 
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I see the thread has built up a life of its own. In any case, I solved the integral after the second post. You need to take the second integral and not. -x=y. You can then add the two integrals together. Then factor out e^x and the answer is 2-2/e.

Also, pka's answer is competent as well.
 
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