how would you evaluate the definite integral of ((1+3x)/x^2)dx between 3 and 1?

- anonymous

how would you evaluate the definite integral of ((1+3x)/x^2)dx between 3 and 1?

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- schrodinger

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- amistre64

split the fraction into two parts with common denominators

- amistre64

1/x^2 + 3x/x^2
(S) x^-2 + 3x^-1 dx | [1,3]

- amistre64

-1/x + 3 ln(x)

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## More answers

- amistre64

-1+0 = -1
(-1/3 + ln(27)) - (-1)

- amistre64

3.9625 there abouts if I did it right :)

- anonymous

that is! thanks :D
can you do this question?
how do u find the antiderivative of sqrt (sinxcosx)? i need to find the volume of the soilid when the given graph is roated around the x-axis.

- amistre64

hmmmm......maybe :)

- anonymous

oh and you need to find the volume between 0 and pie/2

- amistre64

change it to exponent stuff...
(S) sin^(1/2) cos(1/2) dx

- amistre64

let u = sin, du = cos dx might be useful

- anonymous

oh ok. so intergration by parts.

- amistre64

dx = du/cos yeah, im still new at the integrating, but I got some skill :)

- amistre64

u^(1/2) cos^(1/2)
-------- gives us what?
cos

- amistre64

im trying change of variable first :)

- anonymous

actually you know what because i'm finiding the volume, i actually only need to find the antiderivative of sinxcosx

- amistre64

really?...... but I was doing so good :)

- amistre64

u= sin du = cosx dx
(S) u du -> (sin^2)/2

- amistre64

there should be punctuation in there somehere to clear things :)

- amistre64

and if its bounded you dont need the +C

- anonymous

that still doesn't make sense. if you make u=sinx, then du=cos x dx
and dv=cos x and then v=sinx. we're back to where we started...?

- amistre64

if its just the integral of sincos right?

- anonymous

yea. and the equation is uv - integral of v du, when u integrate by parts.

- amistre64

step back to a time when you were first learning this.....
they probably went thru the easier stuff first like u-substitution and change of variables

- amistre64

if we can get it into the form:
(S) u du
then thats all we need to go, no "by parts" or nothing.
(S) sin cos dx
^ ^^^
(S) u du

- amistre64

(S) u du -> (u^2)/2
(sin^2(x))/2 derive this and we get back to sin cos

- anonymous

oh. lol. right. yea i get it...oh man. aha.

- amistre64

you sure you dont need sqrt(sin cos) ??

- anonymous

k, well the question is, find the volume of the solid obtained when the given region y=sqrt (sinxcosx)dx is roatated around the x-axis.

- anonymous

and since the formula for volume is v= pie (s) f(x)^2 dx i can cancel the sqrt right?

- amistre64

which is found by pi (S) [f(x)]^2 dx right?

- amistre64

yes.... sounds good :)

- anonymous

what answer did you get? if you have to find it between pi/2 and 0?

- amistre64

lets see:
sin(x)^2
------ sin(0) = 0 so thats pointless
2
sin (pi/2) = 1 so I get 1/2 as the volume between 0 and 1

- anonymous

i'm lame. my calculator was in degree mode...but i should have been able to figure it out anyways on paper.

- amistre64

lol..... yeah :) these are basic angles and stuff from trig class :)

- anonymous

i hate trig, i just type everything into my calc. ahha.

- amistre64

i made it thru 3 semesters of math without no calculator :)

- amistre64

last test they had odd angles that aint inthe "standard" tables.... broke my streak

- anonymous

ahah. well. thanks a bunch.

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