Question

Really hard calc problem

Answer 1

@AlexandervonHumboldt2
\[\int\limits_{0}^{2} x \sqrt{2x+1}\]

Answer 2

oh i dont know them

Answer 3

you can do this by parts

Answer 4

I understand how to do this until a ceratin point

Answer 5

u=2x+1 AND X=U-1/2

Answer 6

After that i'm lost

Answer 7

\[\frac{ u-1 }{ 2 }\]

Answer 8

if you do it by parts, take u=x, du=dx
dv=sqrt(2x+1) v=(2x+1)^3/2 * 3

Answer 9

So the problem is \[\int\limits_{0}^{2} x \sqrt{2x+1}\] and I got U=2x+1 and x= \[\frac{ u-1 }{ 2 }\]

Answer 10

sub u = 2x+1, du = 2dx\[\frac{ 1 }{ 4 } \int\limits_{1}^{5}(u-1)\sqrt{u}du\] easy from there

Answer 11

my bad, v=(2x+1)^3/2 * 1/3
then uv=uv - int (vdu)

Answer 12

I will show on the \(\normalsize\color{black}{\rm indefinite}\) integral.
\[\Large \int\limits_{ }^{ }x \sqrt{2x+1}~dx\]\[u=2x+1,~~~~~x=\frac{u+1}{2},~~~~\frac{1}{2}du=dx\]\[\frac{1}{4}\Large \int\limits_{ }^{ }u+1 \sqrt{u}~du\]

Answer 13

You don't need parts...

Answer 14

why does the bounds change?

Answer 15

I mean (u+1) like this, in parenthesis

Answer 16

its doable by parts

Answer 17

Because it's a definite integral you have to plug in the bounds in u = 2x+1

Answer 18

where does the 1/4 come from?

Answer 19

\[\frac{1}{4}\Large \int\limits_{ }^{ }(u+1)\sqrt{u}\]\[\frac{1}{4}\Large \int\limits_{ }^{ }(u^{3/2}+u^{1/2})~du\]

Answer 20

and then the power rule... but don;t forget to multiply times 1/4 and the +C.

Answer 21

where did you get the 1/4 though

Answer 22

I mean you need the sums... but you know it.

Answer 23

Ohh, because my substitutions were (u+1)/2 and 91/2) du=dx
so I had to multiply times 1/2 twice

Answer 24

this way I took 1/4 out

Answer 25

I'm sorry I dont understand @SolomonZelman

Answer 26

Do you see why my substitutions were what the were?

Answer 27

no