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

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

Mathematics
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split the fraction into two parts with common denominators
1/x^2 + 3x/x^2 (S) x^-2 + 3x^-1 dx | [1,3]
-1/x + 3 ln(x)

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Other answers:

-1+0 = -1 (-1/3 + ln(27)) - (-1)
3.9625 there abouts if I did it right :)
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.
hmmmm......maybe :)
oh and you need to find the volume between 0 and pie/2
change it to exponent stuff... (S) sin^(1/2) cos(1/2) dx
let u = sin, du = cos dx might be useful
oh ok. so intergration by parts.
dx = du/cos yeah, im still new at the integrating, but I got some skill :)
u^(1/2) cos^(1/2) -------- gives us what? cos
im trying change of variable first :)
actually you know what because i'm finiding the volume, i actually only need to find the antiderivative of sinxcosx
really?...... but I was doing so good :)
u= sin du = cosx dx (S) u du -> (sin^2)/2
there should be punctuation in there somehere to clear things :)
and if its bounded you dont need the +C
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...?
if its just the integral of sincos right?
yea. and the equation is uv - integral of v du, when u integrate by parts.
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
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
(S) u du -> (u^2)/2 (sin^2(x))/2 derive this and we get back to sin cos
oh. lol. right. yea i get it...oh man. aha.
you sure you dont need sqrt(sin cos) ??
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.
and since the formula for volume is v= pie (s) f(x)^2 dx i can cancel the sqrt right?
which is found by pi (S) [f(x)]^2 dx right?
yes.... sounds good :)
what answer did you get? if you have to find it between pi/2 and 0?
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
i'm lame. my calculator was in degree mode...but i should have been able to figure it out anyways on paper.
lol..... yeah :) these are basic angles and stuff from trig class :)
i hate trig, i just type everything into my calc. ahha.
i made it thru 3 semesters of math without no calculator :)
last test they had odd angles that aint inthe "standard" tables.... broke my streak
ahah. well. thanks a bunch.

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