Question

Find the Integral
Integral of (0,2) x/ square root of 1 + 2x^2 dx

Answer 1

did u try substituting \(\large u = 1+2x^2\) ?

Answer 2

\[\int\limits_{0}^{2}\frac{ x }{ \sqrt{1+2x^2} }dx\]

Answer 3

And as @ganeshie8 said, substitute u=1+2x^2
From there, we can get du/dx=4x
So, du/4=xdx

Answer 4

\[\int\limits_{0}^{2}\frac{ 1 }{ \sqrt u }(\frac{ 1 }{ 4 })du\]

Answer 5

can you take it from here?

Answer 6

@helpval22

Answer 7

yes, I substitutes u= 1+2x^2, and du = x^3 dx? ( I think that one is wrong), then 1/2 integral of (0,2) u ^-1/2 du?

Answer 8

no, hold on I think I got where I'm wrong. let me try to solve it, just give me a couple of mins.

Answer 9

alright. take your time :)

Answer 10

Ok, du= 1? so integral of (0,2) 1/ square root of 1 + 2x^2 (1/4) (1) ? Or should I do something else before substituting?

Answer 11

Wait. How did you get du=1?

Answer 12

\[u=1+2x^2\]\[\frac{ du }{ dx }=4x\]\[du=4xdx\]\[\frac{ 1 }{ 4 }du=xdx\]

Answer 13

I didn't know which value was du, maybe 4x or simply xdx? it's what I first assumed..

Answer 14

you just basically differentiate the "substitution equation", which is u=1+2x^2 in this case.
du is not a value; it's the result of differentiation.

Answer 15

Ok, so how do I substitute after that? du = 1/4xdx? I'm a little confused there ( always happens to me in this step) or integral of 0,2) 1/ square root of u (1/4) dx?

Answer 16

watch this

Answer 17

\[\int\limits_{0}^{2}\frac{ x }{ 1+2x^2 }dx=\int\limits_{0}^{2}\frac{ 1 }{ 1+2x^2 }xdx\]

Answer 18

Now make 2 substitutions
u=1+2x^2
du/4=xdx

Answer 19

\[\int\limits_{0}^{2}\frac{ 1 }{ u }(\frac{ 1 }{ 4 })du=\int\limits_{0}^{2}\frac{ 1 }{ 4u }du\]

Answer 20

Correction: there should be square root over both 1+2x^2 and u, but other than that, this is how it's done.

Answer 21

and then after doing that and replacing u values, you integrate the 2 into the x and substract the 0 at the end right?

Answer 22

Yeah.

Answer 23

ok, one last question, should I also substitute for the x in the 4xdx part? so far I am 3/4 (4xdx) -0 . Should I also substitute it for 4(2) dx?

Answer 24

Or just completely ignore the dx and have the answer be 6?

Answer 25

Wait. I forgot something!
When you substitute for an integral, the domain also has to change.
For this case, the domain was [0,2] for x.
So, for u, it would be [1,9]

Answer 26

The final answer should be 1

Answer 27

uhm ok changing that then thank you... but the answer is still 6 ,You don't add or substitute for dx right?

Answer 28

1? How

Answer 29

Since I'm running out of time, I'll post the work.\[\int\limits_{0}^{2}\frac{ x }{ 1+2x^2 }dx \rightarrow \int\limits_{1}^{9}\frac{ 1 }{ 4\sqrt u }du=\left[\frac{ 1 }{ 2 } \sqrt u \right]_{1}^{9}=\frac{ \sqrt 9 }{ 2 }-\frac{ \sqrt 1 }{ 2 }=1\]

Answer 30

Ok, thanks for your help!

Answer 31

yep!