Question

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

Answer 1

split the fraction into two parts with common denominators

Answer 2

1/x^2 + 3x/x^2

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

Answer 3

-1/x + 3 ln(x)

Answer 4

-1+0 = -1

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

Answer 5

3.9625 there abouts if I did it right :)

Answer 6

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.

Answer 7

hmmmm......maybe :)

Answer 8

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

Answer 9

change it to exponent stuff...

(S) sin^(1/2) cos(1/2) dx

Answer 10

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

Answer 11

oh ok. so intergration by parts.

Answer 12

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

Answer 13

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

Answer 14

im trying change of variable first :)

Answer 15

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

Answer 16

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

Answer 17

u= sin du = cosx dx

(S) u du -> (sin^2)/2

Answer 18

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

Answer 19

and if its bounded you dont need the +C

Answer 20

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...?

Answer 21

if its just the integral of sincos right?

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

(S) u du -> (u^2)/2

(sin^2(x))/2 derive this and we get back to sin cos

Answer 26

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

Answer 27

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

Answer 28

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.

Answer 29

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

Answer 30

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

Answer 31

yes.... sounds good :)

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

ahah. well. thanks a bunch.