Question

integrate x(x-1)^(1/2)
lower bound = 1
upper bound = 2

I know the answer but when I try and do it out I keep getting the wrong one, any help?

Answer 1

\[\int\limits_{1}^{2}x \sqrt{x-1} dx\]
Did you do u=x-1 => du =dx
Also u=x-1 => u+1=x

And if x=1 then u=1-1=0
And if x=2 then u=2-1=1

So we have
\[\int\limits_{0}^{1} (u+1) \sqrt{u} du\]

Answer 2

Did you get this far?

Answer 3

oh and i like robots :)

Answer 4

myininaya. i need your help. I messaged you. but you didn't reply me. :(

Answer 5

i cant see what you got because it says math processing answer

Answer 6

Try refreshing...

Answer 7

@imalittletoyrobotfromthefuture did that work?

Answer 8

yea it worked, i have that now. I didnt know what to do because I didnt recognize that x = u + 1

Answer 9

Well if u=x-1
then u+1=x (I just added 1 on both sides for u=x-1)

Answer 10

yes... haha i got that, what i had before though was:

x\[\int\limits_{1}^{2}x \sqrt{u}du\]

Answer 11

But your integral needs to be in terms of one variable and when you make the switch from x to u your limits also need to switch from x to u

Answer 12

yes, thats why i was having trouble, i didnt recognize the fact that I could put x in terms of u, now that I have that though I'm unsure of what to do next

Answer 13

\[\int\limits\limits_{0}^{1} (u+1) \sqrt{u} du \]
So this is what I wrote above

Answer 14

Do you understand exactly how I got it?

Answer 15

yes, what next though? this?:

\[\int\limits_{0}^{1}((u^2/2)+u)((2u^3/2)/3)\]

Answer 16

?

Answer 17

isn't that the anti-derivitave of the equation \[\int\limits_{0}^{1}(u+1)\sqrt{u}du\]?

Answer 18

\[\int\limits_{0}^{1}(u+1) \sqrt{u} du=\int\limits_{0}^{1}(u+1)u^\frac{1}{2} du=\int\limits_{0}^{1}(u \cdot u^\frac{1}{2}+1 \cdot u^\frac{1}{2}) du\]
\[=\int\limits_{0}^{1}(u^\frac{3}{2}+u^\frac{1}{2}) du\]

Answer 19

Now we integrate...

Answer 20

\[[\frac{u^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}]_0^1\]

Answer 21

\[=[\frac{1^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\frac{1^{\frac{1}{2}+1}}{\frac{1}{2}+1}]-[\frac{0^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\frac{0^{\frac{1}{2}+1}}{\frac{1}{2}+1}]\]

Answer 22

robot you have questions?

Answer 23

16/15 :) thanks for the explanation, very helpful and yes, robots are sick. check this out http://www.cnn.com/2012/03/04/opinion/ted-kumar-flying-robots/index.html?iref=obnetwork

Answer 24

lol