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Integration of expo

  • Thread starter jwxie
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  • #1
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Homework Statement



[1] integral of e^(-ln(x))
[2] integral of 2/(sqrt(x)) * e^(sqrt(x)))

Homework Equations



The Attempt at a Solution



Is [1] the same as writing 1/e^lnx?

In general, example of 1 and 2, we treat the negative of an expo power, e^-x as 1/e^x right?

So for [2], u = sqrt(x) (from the e), should be written as -sqrt(x)?

I thought in regular integration:
x^2 / x^3 , where u is x^3, and it is not equivalent as -x^3, or x^-3
 

Answers and Replies

  • #2
33,162
4,847

Homework Statement



[1] integral of e^(-ln(x))
[2] integral of 2/(sqrt(x)) * e^(sqrt(x)))

Homework Equations



The Attempt at a Solution



Is [1] the same as writing 1/e^lnx?
Yes, but you might as well take it a step further, since 1/e^(lnx) = 1/x, when x > 0.
In general, example of 1 and 2, we treat the negative of an expo power, e^-x as 1/e^x right?

So for [2], u = sqrt(x) (from the e), should be written as -sqrt(x)?
Are you asking whether e^(sqrt(x)) = sqrt(x)? No, it's not. If that' not what you're asking, please clarify
I thought in regular integration:
x^2 / x^3 , where u is x^3, and it is not equivalent as -x^3, or x^-3
 
  • #3
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Hi, thanks. For [2] I am sorry, I wasn't clear.

There was a typo in [2]
here is the right one, with answer from wolfram alpha
http://www.wolframalpha.com/input/?i=integral+of++2%2F%28sqrt%28x%29+*+e^%28sqrt%28x%29%29%29

Initially I took sqrt(x) as my u variable (from e), and du = 1/2 sqrt(x)^-1/2 dx

This would never get a negative, so I turned to our previous agreement that, e^-x = 1/e^x

This is being done in wolfram alpha.... And usually I avoid taking - exponent whenever possible...

I thought in general, a / b^x, where b^x = b^-x, I can ignore the minus in u substitution

So can you show me how to solve [2] without using the method (take -exponent), and go straight with the original format?
 
  • #4
ideasrule
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b^x does not equal b^-x; 1/b^x equals b^-x. You can't randomly ignore negative signs.

You have u and you have du, so just rewrite everything in the original integral in terms of u and du and integrate the result.
 
  • #5
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e^-x = 1/e^x

now if you check the link i provided, you will see wolfram alpha show a step that i don't understand.

click on that link, and click on show steps

WA rewrites the original function to e^-sqrt(x)/sqrt(x), so that u = -sqrt(x)

but as i said, usually i do it straightforward, so that in general, a/b^x, i don't care 1/b^x = b^-x, i can just put u = b^x
for example
2/2x, can be rewritten as 2( 2*x^-1)./ However, i don't have to worry about the minus sign in 2/2x form
i got stright to 2/2x, u = 2x, du = 2 dx, so 1/2du = dx

compare this to what i just said about wolfram, you will see my point
 
  • #6
33,162
4,847
2/2x, can be rewritten as 2( 2*x^-1)./
Not true. 2(2*x-1) = 4/x [itex]\neq[/itex]2/(2x)

Simplify it first. 2/(2x) = 1/x = x-1
 
  • #7
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Not true. 2(2*x-1) = 4/x [itex]\neq[/itex]2/(2x)

Simplify it first. 2/(2x) = 1/x = x-1
Hi. Thanks. I am such a fool thatI get these simple things wrong.

Yes you are right.
1/x = x^-1
So let's stick with this one

I would use 1/x, where u = x, and du = 1 dx
This would not require negative exponent at all

So in my attempt

(2/sqrt(x) * e^(sqrt(x)) <=----- the original function
u = sqrt(x)
du = 1/2 sqrt-1(x) dx
solve for dx, I get 2 sqrt(x) du = dx, and this i can replace dx by that expression

i get 2*sqrt(x) * 2 / sqrt(x) e^u
this produce 4/e^u
integral of 4/e^u, take out the 4, i am left with 1/e^u, but this cannot be solve unless we turn it back to -1 power, so integral of e^-u gives -1/1 * e^-u, and so the final answer is -4e^-u

now i get it.
thanks.
 
Last edited:
  • #8
33,162
4,847
(2/sqrt(x) * e^(sqrt(x)) <=----- the original function
u = sqrt(x)
du = 1/2 sqrt-1(x) dx
solve for dx, I get 2 sqrt(x) du = dx, and this i can replace dx by that expression

i get 2*sqrt(x) * 2 / sqrt(x) e^u
this produce 4/e^u
integral of 4/e^u, take out the 4, i am left with 1/e^u, but this cannot be solve unless we turn it back to -1 power, so integral of e^-u gives -1/1 * e^-u, and so the final answer is -4e^-u

now i get it.
I don't think you do. For one thing, your final answer should not be in terms of x. You need to "undo" your substitution. For another thing, the antiderivative you got is not correct.

You are making this much more complicated than it really. For this integral -
[tex]\int \frac{2}{\sqrt{x}} e^{\sqrt{x}} dx[/tex]

You have the right substitution, u = sqrt(x), so du = dx/(2sqrt(x)). Notice that the integral has almost what you need for du.
 
  • #9
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I don't think you do. For one thing, your final answer should not be in terms of x. You need to "undo" your substitution. For another thing, the antiderivative you got is not correct.

You are making this much more complicated than it really. For this integral -
[tex]\int \frac{2}{\sqrt{x}} e^{\sqrt{x}} dx[/tex]

You have the right substitution, u = sqrt(x), so du = dx/(2sqrt(x)). Notice that the integral has almost what you need for du.
Sorry, I mean this integral...
gossh I should be more considerate next time by putting it in latex form, lol
[tex]\int \frac{2}{{\sqrt{x}} e^{\sqrt{x}}}dx[/tex]
 
  • #10
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I attempted the problem using a u - sub. Bare with me, my solution may not be correct, however here is my idea. I is integral sign

sub u = sqrt(x) .. The integral becomes 4 I u / u e^ u du = 4 I e ^ - u du. = 4 ( sinh( u) - cosh (u )) + K. Sub back and finish up.

EDIt:: That should be - 4 I e ^ -u du = - 4 ( sinh ( u ) - cosh ( u ) ) + K.
 
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