Question

Definite Integral
Can I do this by substitution?

Answer 1

\[\int\limits_{1}^{5}\frac{ x }{ \sqrt{2x-1} }dx\]

Answer 2

I don't think I can use substitution to ingrate this one but my directions are to do so.

Answer 3

I can't use integral by parts since it is not taught in this course

Answer 4

I can just integrate using my calculator

Answer 5

You could use \(u=2x-1\) so \(du=2 dx\)
Solve for x to get: \(\Large x=\frac{u+1}{2}\)

@ganeshie8 ?

Answer 6

thats a good idea ! also u may try u = \(\sqrt{2x-1}\) also, not sure...

Answer 7

\(du = \dfrac{2dx}{\sqrt{2x-1}} \implies \dfrac{dx}{\sqrt{2x-1}} = \dfrac{du}{2}\)

Answer 8

the indefinite integral becomes

\[ \large \dfrac{1}{2}\int u^2 + 1~ du \]

Answer 9

1/4 *

Answer 10

You could also plug in the x-limits into the u-sub equation to get the integral in terms of u instead of going back and sub x.

Answer 11

Lol lui XD

Answer 12

Batman >.>

Answer 13

You can use u = 2x-1 du = 2dx

\[\frac{ 1 }{ 2 } \int\limits \frac{ u+1 }{ 2\sqrt{u} }du\]

it's what lui recommended, this could work, but will take a while, you'll have to expand the integrand.

Answer 14

Actually no, this is pretty good

Answer 15

\[\Large \frac{1}{4} \int\frac{x+1}{\sqrt{u}}~du \]
^Simplier

Answer 16

omg isn't there an easy method to do this instead of substitution

Answer 17

Force it into a trig sub?

Answer 18

assuming the sky is the limit and I could use any valid method other than substitution

Answer 19

\[\frac{ 1 }{ 2 } \int\limits (\frac{ \sqrt{u} }{ 2 }+\frac{ 1 }{ 2\sqrt{u} }) du = \frac{ 1 }{ 4 } \int\limits \frac{ 1 }{ \sqrt{u} }du +\frac{ 1 }{ 4 }\int\limits \sqrt{u}du\]

Answer 20

My knowledge in calc is limited, sorry :P

Ask batman~

Answer 21

Well I'd say this is the best way tbh, you can try other stuff, but might be a bit tedious, idk.

Answer 22

@iambatman are you using integration by parts

Answer 23

No, just substitution, and then algebra :)

Answer 24

Only the start of the integration is calculus, then it goes back to high school math, remember the rules :P

Answer 25

You COULD use IBP but it'll be harder if you ask me..

Answer 26

Are you allowed to use IBP?

Answer 27

substitution is always the easiest method,
integration by parts is always the hardest method

NON-NEGOTIABLE

Answer 28

Agreed

Answer 29

no but I have taken Cal I, II, III a very long time ago

Answer 30

I think substitution is the best way to approach this integral, well that's what the first thing comes in mind, then if that doesn't work you start testing other methods.

Answer 31

ok thanks I will come back to these problems later

Answer 32

Np, try doing this problem again, without help ;P, instead of doing definite integrals, try just focusing on how you would approach them as indefinite.

Answer 33

Because, lets be honest once you do the integration part, putting the limits is easy :)