Integration problem (1 Viewer)

spidey4 · Nov 16, 2009

Hi can anybody help to integrate (3x-2)(2x-3)^1/2

Thanks

addikaye03 · Nov 16, 2009

spidey4 said:
Hi can anybody help to integrate (3x-2)(2x-3)^1/2

Thanks

Are you sure that's right? I can't get it using only 3U knowledge

Drongoski · Nov 16, 2009

addikaye03 · Nov 16, 2009

Drongoski said:
$\100dpi Is\ it\ \ i) \int \sqrt{(3x-2)(2x-3)}\ dx\ \ \ or\ \ \ ii)\ \int (3x-2)\sqrt{2x-3}\ dx\ \ \ ??$

i think it's (3x-2)[(2x-3)^-1/2]

that would make sense, but don't think it would be 3U

Either of your above is pretty ugly to solve, especially i) lol

Trebla · Nov 16, 2009

If it was (ii), ∫ (3x - 2)(2x - 3)^0.5 dx
Let u = 2x - 3
=> x = (u + 3)/2
=> dx = du/2
Hence
3x - 2 = 3(u + 3)/2 - 2
= (3u + 5)/2

∫ (3x - 2)(2x - 3)^0.5 dx = (1/4) ∫ (3u + 5)u^0.5 du
= (1/4) ∫ (3u^1.5 + 5u^0.5) du
= (1/4) [6u^2.5/5 + 10u^1.5/3] + c
= (1/4) [6(2x - 3)^2.5/5 + 10(2x - 3)^1.5/3] + c

Hope the working's right....lol

addikaye03 · Nov 16, 2009

Trebla said:
If it was (ii), ∫ (3x - 2)(2x - 3)^0.5 dx
Let u = 2x - 3
=> x = (u + 3)/2
=> dx = du/2
Hence
3x - 2 = 3(u + 3)/2 - 2
= (3u + 5)/2

∫ (3x - 2)(2x - 3)^0.5 dx = (1/4) ∫ (3u + 5)u^0.5 du
= (1/4) ∫ (3u^1.5 + 5u^0.5) du
= (1/4) [6u^2.5/5 + 10u^1.5/3] + c
= (1/4) [6(2x - 3)^2.5/5 + 10(2x - 3)^1.5/3] + c

Hope the working's right....lol

Wolfram Mathematica Online Integrator

How did they do that? Is it something simple that i'm missing?

Trebla · Nov 16, 2009

Note that:
(1/4) [6(2x - 3)^2.5/5 + 10(2x - 3)^1.5/3] + c
= (1/4) (2x - 3)^1.5 [6(2x - 3)/5 + 10/3] + c
= (1/4) (2x - 3)^1.5 [(36x - 54 + 50)/15] + c
= (1/4) (2x - 3)^1.5 [(36x - 4)/15] + c
= (1/4) (4/15) (2x - 3)^1.5 (9x - 1) + c
= (1/15) (2x - 3)^1.5 (9x - 1) + c

shady145 · Nov 16, 2009

Trebla said:
If it was (ii), ∫ (3x - 2)(2x - 3)^0.5 dx
Let u = 2x - 3
=> x = (u + 3)/2
=> dx = du/2
Hence
3x - 2 = 3(u + 3)/2 - 2
= (3u + 5)/2

∫ (3x - 2)(2x - 3)^0.5 dx = (1/4) ∫ (3u + 5)u^0.5 du
= (1/4) ∫ (3u^1.5 + 5u^0.5) du
= (1/4) [6u^2.5/5 + 10u^1.5/3] + c
= (1/4) [6(2x - 3)^2.5/5 + 10(2x - 3)^1.5/3] + c

Hope the working's right....lol

from memory this is 4u method where you have to make your own substitution?
adikaye i thought this was your original problem, is there a way to do this 3u style, or is making your own substitution 3u? =S

anGchan · Nov 16, 2009

im pretty sure 'own substitution' is not limited to the realm of 4u

spidey4 · Nov 16, 2009

Hi sorry the original problem should be

Integral of (3x-2)(2x-3)^-0.5

The answer is (x+1)(2x-3)^0.5 + C

can you please show the steps involved to get this.

Thank you very much, and sorry about my first blunder.

Trebla · Nov 17, 2009

spidey4 said:
Hi sorry the original problem should be

Integral of (3x-2)(2x-3)^-0.5

The answer is (x+1)(2x-3)^0.5 + C

can you please show the steps involved to get this.

Thank you very much, and sorry about my first blunder.

It's pretty much the same steps as I outlined in the above posts.

The Nomad · Nov 17, 2009

Answer is attached.

untouchablecuz · Nov 17, 2009

$\int \frac{3x-2}{\sqrt{2x-3}}dx\\=\int (\frac{3x}{{\sqrt{2x-3}}}-2(2x-3)^{-\frac{1}{2}})dx\\=\int (\frac{3}{2}\times\frac{2x-3+3}{{\sqrt{2x-3}}}-2(2x-3)^{-\frac{1}{2}})dx\\= \frac{3}{2}\int \frac{2x-3+3}{{\sqrt{2x-3}}}-\int 2(2x-3)^{-\frac{1}{2}}dx\\= \frac{3}{2}\int (\frac{2x-3}{{\sqrt{2x-3}}}+\frac{3}{{\sqrt{2x-3}}})dx-\int 2(2x-3)^{-\frac{1}{2}}dx\\= \frac{3}{2}\int ((2x-3)^{\frac{1}{2}}+\frac{3}{{\sqrt{2x-3}}})dx-\int 2(2x-3)^{-\frac{1}{2}}dx\\= \frac{3}{2}\int ((2x-3)^{\frac{1}{2}}+\frac{3}{2}(2)(2x-3)^{-\frac{1}{2}})dx-\int 2(2x-3)^{-\frac{1}{2}}dx\\= \frac{3}{2}\int (2x-3)^{\frac{1}{2}}dx+\frac{9}{4}\int2(2x-3)^{-\frac{1}{2}}dx-\int 2(2x-3)^{-\frac{1}{2}}dx\\=\frac{3}{2}\int (2x-3)^{\frac{1}{2}}dx+(\frac{9}{4}-1)\int2(2x-3)^{-\frac{1}{2}}dx$

which you can integrate

edit: substitution is much faster

spidey4 · Nov 17, 2009

Thank you so much

addikaye03 · Nov 17, 2009

Trebla said:
Note that:
(1/4) [6(2x - 3)^2.5/5 + 10(2x - 3)^1.5/3] + c
= (1/4) (2x - 3)^1.5 [6(2x - 3)/5 + 10/3] + c
= (1/4) (2x - 3)^1.5 [(36x - 54 + 50)/15] + c
= (1/4) (2x - 3)^1.5 [(36x - 4)/15] + c
= (1/4) (4/15) (2x - 3)^1.5 (9x - 1) + c
= (1/15) (2x - 3)^1.5 (9x - 1) + c

Ahh ofcourse, thanks for the clarification Treb

shady145 said:
from memory this is 4u method where you have to make your own substitution?
adikaye i thought this was your original problem, is there a way to do this 3u style, or is making your own substitution 3u? =S

Yeah, that was my Q, the original Q couldn't have been done using 3U (since no substitution given is 4U, except for this Q: int. sin^2@cos@d@). But as you can see the Q was written incorrect, solutions provided by Unthouchablecuz and Nomad would be the answer for the fixed question, and would be within 3U. I still think it's an ugly Q for 3U though, 4U it's pretty standard though

Ben1220 · Nov 17, 2009

sikhman said:
no i was told thinking up your own substitution was only in 4U, in 3U, they tell you which one to use (and sometimes they do in 4U)

even so, its not that hard anyway to do a linear substitution, all you do is substitute u = (whatever is within the function that isn't linear), and then rewrite the linear part so that it is with respect to u.

The Nomad · Nov 17, 2009

sikhman said:
integration in general is not hard

What a useful post, considering how much it helped to solve the question.

Integration problem (1 Viewer)

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