Question

can someone please help me try and figure out where I'm going wrong with this definite integral? :)

Answer 1

\[2\pi \int\limits_{0}^{4}x \sqrt(1+\frac{ x^2 }{ 16 }) dx\]

Answer 2

Looks like an easy u sub.

Answer 3

this was what I did: let u=1 + (X^2)16, therefore du= x/8 dx and dx=8/x du

Answer 4

whoops typo divided by 16

Answer 5

so then I got \[16\pi \int\limits_{0}^{4}\sqrt(u)du\]

Answer 6

is that right so far?

Answer 7

You should change the limits of integration.

Answer 8

\[
(0,1)\to (u(0),u(1))
\]

Answer 9

I can't carry it through to the end? That's the way we've done it in class. Shouldn't it give the same answer eventually?

Answer 10

so for example I got \[\frac{ 2 }{ 3 }u ^{3/2} \] from 0 to 4, and then... finally I got

Answer 11

\[\frac{ 2 }{ 3 }(1+\frac{ x^2 }{ 16 })\] evaluated from 0 to 4, and then that answer multiplied by \[16 \pi\]

Answer 12

You're not getting it...

Answer 13

When you do u sub, the integration limits change

Answer 14

you can keep the same limits if you substitute whatever you substituted back in at the end, which I did though

Answer 15

you cannot integrate with respect to u and use limits defined for x

Answer 16

whatever expression you used to define your u, plug in x limits to get u limits

Answer 17

yeah, but I plugged the \[1+\frac{ x^2 }{ 16 }\] back in at the end before I did my calculations, so I can't keep the limits?

Answer 18

if x limits are 0 and 4, then for u(x)=1+x^2/16
u(0)=1+(0)^2/16=1
u(4)=1+(4)^2/16=2

Answer 19

okay, just a sec let me see if I get the right answer using that :)

Answer 20

your new limits of integration with respect to u are now 1 and 2, once you make the u substitution you can forget about x since you redifined your variable and therefore you need to redifine your limits

Answer 21

I did. Thanks :)

Answer 22

cool

Answer 23

make sure you understand this, because it will be on a test, 100% guaranteed