# u subs and x bounds on integrals

• Apr 5th 2009, 11:53 AM
squid
u subs and x bounds on integrals
Could someone please explain how to use u subs when taking an integral? We learned this awhile ago and I never completely understood. Here is an example of a problem where, I think, I should be using a u sub:
ack! I can't figure out how to type this in........ sooooo....

The integral of x squared minus 1 over x between the bounds of e and 1

Hope that makes sense.
Thank you :)

Also, I know I'm only supposed to ask one question, but how would the answer change if the upper bound were x? Would I just plug in x to the final solution, or is there a special rule?
• Apr 5th 2009, 12:11 PM
stapel
If I understand you correctly, you're asking what sort of substitution might be helpful in the following:

. . . . . $\int_1^e\, \frac{x^2\, -\, 1}{x}\, dx$

It should be noted that there are oftentimes many ways of doing an integral. No one is more "right" than another. Since this can be simplified, you could work with the following:

. . . . . $\int_1^e\, x\, -\, \frac{1}{x}\, dx$

In this way, no substitution would be necessary. But you could fiddle around with substitutions, and notice that:

. . . . . $\mbox{for }\, \ln(x)\, =\, u,\, \mbox{ we have }\, \frac{1}{x}\, dx\, =\, du$

Then:

. . . . . $x^2\, -\, 1\, =\, e^{2u}\, -\, 1$

Does that help at all...? (Wink)
• Apr 5th 2009, 12:14 PM
skeeter
Quote:

Originally Posted by squid
Could someone please explain how to use u subs when taking an integral? We learned this awhile ago and I never completely understood. Here is an example of a problem where, I think, I should be using a u sub:
ack! I can't figure out how to type this in........ sooooo....

The integral of x squared minus 1 over x between the bounds of e and 1

Hope that makes sense.
Thank you :)

Also, I know I'm only supposed to ask one question, but how would the answer change if the upper bound were x? Would I just plug in x to the final solution, or is there a special rule?

I'm assuming 1 is the lower limit of integration and e is the upper.

now ... is it

$\int_1^e \frac{x^2-1}{x} dx$

or

$\int_1^e x^2 - \frac{1}{x} \, dx$

???

fyi, no substitution required in either case.

$\int_1^e x - \frac{1}{x} \, dx$

$\left[\frac{x^2}{2} - \ln{x}\right]_1^e = \left(\frac{e^2}{2} - 1\right) - \left(\frac{1}{2} - 0 \right) = \frac{e^2-3}{2}$

or ...

$\int_1^e x^2 - \frac{1}{x} \, dx$

$\left[\frac{x^3}{3} - \ln{x}\right]_1^e = \left(\frac{e^3}{3} - 1\right) - \left(\frac{1}{3} - 0\right) = \frac{e^3-4}{3}$
• Apr 5th 2009, 12:25 PM
squid
haha I see now, thank you both. And I meant the equation the way staple wrote it, or skeeter's first one.

Would you also mind answering if there is a special way to solve an integral with an upper bound of x? For example, if the equation I used above had an upper bound of x rather than e, would I just plug in x at the end, or do I do something different?
• Apr 5th 2009, 12:32 PM
skeeter
Quote:

Originally Posted by squid
haha I see now, thank you both. And I meant the equation the way staple wrote it, or skeeter's first one.

Would you also mind answering if there is a special way to solve an integral with an upper bound of x? For example, if the equation I used above had an upper bound of x rather than e, would I just plug in x at the end, or do I do something different?

$\int_1^x \frac{t^2 - 1}{t} \, dt$

t is a dummy variable, just so that it's different from $x$.

the value of the definite integral is now a function of $x$.
• Apr 5th 2009, 12:36 PM
squid
so would that mean that my final answer would be (x squared over 2) minus lnx minus 1/2?
• Apr 5th 2009, 12:44 PM
skeeter
Quote:

Originally Posted by squid
so would that mean that my final answer would be (x squared over 2) minus lnx minus 1/2?

yes, and that would be a function.