Definite integral

  • Thread starter zeno23
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Homework Statement


Evaluate the definite integral.


Homework Equations


[tex]
\int_{1}^{2} ( 2e^{-4x} -\frac{1}{x^2} ) dx
[/tex]

Answer given by the book: [tex]\frac{1}{2}(e^{-4}-e^{-8}-1)[/tex]

The Attempt at a Solution



u = 4x; x = u/4; du = 4 dx; dx = du/4;

[tex]\frac{1}{2}\int e^{-u} du - 4\int u^{-2} du[/tex]

[tex]\frac{1}{2}\int_{1}^{2} e^{-4x} dx - 4\int_{1}^{2} \frac{1}{16x^2} dx[/tex]

... and I continue down this path to wind up with:

[tex] \frac{1}{2e^8}-\frac{1}{2e^4}-\frac{21}{32}[/tex]

In other words, I've apparently no idea what I'm doing, and would appreciate any help.
 

Answers and Replies

  • #2
35,227
7,048

Homework Statement


Evaluate the definite integral.


Homework Equations


[tex]
\int_{1}^{2} ( 2e^{-4x} -\frac{1}{x^2} ) dx
[/tex]

Answer given by the book: [tex]\frac{1}{2}(e^{-4}-e^{-8}-1)[/tex]

The Attempt at a Solution



u = 4x; x = u/4; du = 4 dx; dx = du/4;

[tex]\frac{1}{2}\int e^{-u} du - 4\int u^{-2} du[/tex]

[tex]\frac{1}{2}\int_{1}^{2} e^{-4x} dx - 4\int_{1}^{2} \frac{1}{16x^2} dx[/tex]

... and I continue down this path to wind up with:

[tex] \frac{1}{2e^8}-\frac{1}{2e^4}-\frac{21}{32}[/tex]

In other words, I've apparently no idea what I'm doing, and would appreciate any help.

Split the integral into two integrals:
[tex]\int_{1}^{2} 2e^{-4x} dx - \int_{1}^2 x^{-2}} dx[/tex]

It looks like you are letting your substitution slop over into the 2nd part of your integral.

For your substitution use u = -4x, so du = -4 dx.

After you have found an antiderivative for the 1st integral, undo your substitution and evaluate the antiderivative at x = 2 and x = 1.

Alternatively, you can write new limits of integration for the 1st integral by converting the two values of x into values of u.
 
  • #3
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Wait, I got it. It should just be e^u, so e^(-4*2). /facepalm

Thanks again.

----------------

Thanks for the response. I've managed to arrive at the proper answer, but I don't grasp the logic completely. I start with:

[tex] \int_{1}^{2} 2e^{-4x}[/tex]

I set u=-4x and dx=(-1/4) du, so then:
[tex] - \frac{1}{2} \int e^u du = -\frac{ e^{ \frac{u^2}{2} } }{2}[/tex].

Putting this back into terms of X, I get: [tex] -\frac{e^{\frac{(-4x)^2}{2}}}{2} [/tex]

But I it looks like it should be [tex] -\frac{e^{-4\frac{x^2}{2}}}{2} [/tex]. When I use this, I get the proper answer. Since u=(-4x), why do we multiply x^2 by -4 in the final step rather than (-4x)^2 ?
 
Last edited:
  • #4
35,227
7,048
Your antiderivative is wrong.
[tex]\int e^u du = e^u + C[/tex]
 
  • #5
35,227
7,048
BTW, in your original work, you used the same substitution on the 2nd part of your integral, which is the wrong thing to do.
 

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