Help with an integration problem. watch

Darke1231 · 07-02-2018 18:35

Attached below an integration problem, giving me a hard time. Could someone solve it step by step and in details.https://cdn.discordapp.com/attachmen...25/unknown.png

BobbJo · 07-02-2018 18:37

Full solutions are not allowed.

Try the substitution as the numerator is a constant multiple of the derivative of the denominator.

Darke1231 · 07-02-2018 19:32

(Original post by BobbJo)
Full solutions are not allowed.

Try the substitution as the numerator is a constant multiple of the derivative of the denominator.

As, I mentioned ; I did that already here's what I ended up with.

Hence,that's why am asking for the solution

ghostwalker · 07-02-2018 20:12

(Original post by Darke1231)
As, I mentioned ; I did that already here's what I ended up with.

Hence,that's why am asking for the solution

I make it $\displaystyle \frac{1}{4a}\ln\left(\frac{17}{2 }\right)$

As BobbJo said, posting full solutions is against forum rules. Post your working if you'd like someone to check it.

Darke1231 · 07-02-2018 20:24

Here's how I approached it.
Find $\displaystyle \int_{a}^{2a}\[\frac{x^3}{a^5+ax^4}$
let
Therefore,
, = $\frac{1}{4ax^3}$
, = $\frac{1}{4a}$
We will define the boundaries later, integrating:
$\displaystyle \int\frac{x^3}{u}$
Since = $\frac{1}{4a}$
$\displaystyle \int\frac{1}{u}$ * $\frac{1}{4a}$
Integrating we get;
* $\frac{u}{4a}$
Re-defining the limits as we are dealing with ;
Since
Where =a, =
,Where =2a, =
So your definite integral will look like this: $\displaystyle \int_{2a^5}^{17a^5}$
Subtracting " $\mathrm{After \ substantiating \ in \ value \ got \ from \ the\ integration }$ " the upper limit from the lower limit will result in $\frac{15a^4}{4}$ * $ln\frac{17}{2}$
However, the answer I got is incorrect , point out the mistake.
Thanks.

Darke1231 · 07-02-2018 20:27

(Original post by ghostwalker)
I make it $\displaystyle \frac{1}{4a}\ln\left(\frac{17}{2 }\right)$

As BobbJo said, posting full solutions is against forum rules. Post your working if you'd like someone to check it.

My answer is wating approval from a mod.

ghostwalker · 07-02-2018 21:38

(Original post by Darke1231)
Here's how I approached it.
Find $\displaystyle \int_{a}^{2a}\[\frac{x^3}{a^5+ax^4}$
let
Therefore,
, = $\frac{1}{4ax^3}$
, = $\frac{1}{4a}$
We will define the boundaries later, integrating:
$\displaystyle \int\frac{x^3}{u}$
Since = $\frac{1}{4a}$
$\displaystyle \int\frac{1}{u}$ * $\frac{1}{4a}$
Integrating we get;
* $\frac{u}{4a}$

You seem to have acquired an extra "u".

Should be $\displaystyle\ln |u|\times \frac{1}{4a}$

Re-defining the limits as we are dealing with ;
Since
Where =a, =
,Where =2a, =
So your definite integral will look like this: $\displaystyle \int_{2a^5}^{17a^5}$
Subtracting " $\mathrm{After \ substantiating \ in \ value \ got \ from \ the\ integration }$ " the upper limit from the lower limit will result in $\frac{15a^4}{4}$ * $ln\frac{17}{2}$
However, the answer I got is incorrect , point out the mistake.
Thanks.

Aside from going wrong earlier, if you had continued along this path, you'd have ended up with:

$\displaystyle \frac{17a^4}{4}\ln 17 - \frac{2a^4}{4}\ln 2$ which doesn't simplify to what you put.

K-Man_PhysCheM · 07-02-2018 21:39

(Original post by Darke1231)
$\displaystyle \int\frac{1}{u}$ * $\frac{1}{4a}$
Integrating we get;
* $\frac{u}{4a}$

Thanks.

The integration there is incorrect. $\displaystyle \int \dfrac{\mathrm{d}u}{4au} = \frac{1}{4a}\ln|u|$

I haven't worked through the rest of the question, but does that solve your problem?

Darke1231 · 08-02-2018 02:08

(Original post by ghostwalker)
You seem to have acquired an extra "u".

Should be $\displaystyle\ln |u|\times \frac{1}{4a}$

Aside from going wrong earlier, if you had continued along this path, you'd have ended up with:

$\displaystyle \frac{17a^4}{4}\ln 17 - \frac{2a^4}{4}\ln 2$ which doesn't simplify to what you put.

I see, Thanks. However, in integrating:
$\displaystyle\int\ \frac{1}{u}*\frac{1}{4a}$
Why multypling them then integrating? and why not integrating $\frac{1}{u}$ then integrating $\frac{1}{4a}$ as a constant? to be $\frac{u}{4a}$ As this what I did (How I ended up with the extra )

Darke1231 · 08-02-2018 02:13

(Original post by K-Man_PhysCheM)
The integration there is incorrect. $\displaystyle \int \dfrac{\mathrm{d}u}{4au} = \frac{1}{4a}\ln|u|$

I haven't worked through the rest of the question, but does that solve your problem?

I see, Thanks. However, in integrating:
$\displaystyle\int\ \frac{1}{u}*\frac{1}{4a}$
Why multypling them then integrating? and why not integrating $\frac{1}{u}$ then integrating $\frac{1}{4a}$ as a constant? to be $\frac{u}{4a}$ As this what I did (How I ended up with the extra )

K-Man_PhysCheM · 08-02-2018 07:47

(Original post by Darke1231)
I see, Thanks. However, in integrating:
$\displaystyle\int\ \frac{1}{u}*\frac{1}{4a}$
Why multypling them then integrating? and why not integrating $\frac{1}{u}$ then integrating $\frac{1}{4a}$ as a constant? to be $\frac{u}{4a}$ As this what I did (How I ended up with the extra )

When you integrate a function multiplied by a constant, you just integrate the function itself and leave the constant multiple unaffected.

eg $\displaystyle \int 5a \times u \ \mathrm{d}u = 5a \times \frac{1}{2}u^2 = \frac{5}{2}au + k$

where and are constants.

This example is no different.

ghostwalker · 08-02-2018 07:53

(Original post by Darke1231)
I see, Thanks. However, in integrating:
$\displaystyle\int\ \frac{1}{u}*\frac{1}{4a}$
Why multypling them then integrating? and why not integrating $\frac{1}{u}$ then integrating $\frac{1}{4a}$ as a constant? to be $\frac{u}{4a}$ As this what I did (How I ended up with the extra )

Well, 1/4a is just a constant, you can pull it out of the integral.

$\displaystyle \int \frac{1}{4a}\times \frac{1}{u}\; du = \frac{1}{4a}\int\frac{1}{u}\; du$

If you're going to try treat it as a function of u, then you're trying to integrate a product of two functions, in which case you need to use integration by parts. It's massive overkill, but you might find it useful as an exercise to do it that way.

There is no justfitication for integating the product of two functions one at a time and multiplying them. If there were you could have, for example:

$\displaystyle\int x^2\;dx = \int x \times x \;dx = \frac{x^2}{2}\times\frac{x^2}{2} = \frac{x^4}{4}$ which is clearly WRONG.

Darke1231 · 08-02-2018 08:08

(Original post by ghostwalker)
Well, 1/4a is just a constant, you can pull it out of the integral.

$\displaystyle \int \frac{1}{4a}\times \frac{1}{u}\; du = \frac{1}{4a}\int\frac{1}{u}\; du$

If you're going to try treat it as a function of u, then you're trying to integrate a product of two functions, in which case you need to use integration by parts. It's massive overkill, but you might find it useful as an exercise to do it that way.

There is no justfitication for integating the product of two functions one at a time and multiplying them. If there were you could have, for example:

$\displaystyle\int x^2\;dx = \int x \times x \;dx = \frac{x^2}{2}\times\frac{x^2}{2} = \frac{x^4}{4}$ which is clearly WRONG.

I see, Thanks.

Darke1231 · 08-02-2018 08:10

(Original post by K-Man_PhysCheM)
When you integrate a function multiplied by a constant, you just integrate the function itself and leave the constant multiple unaffected.

eg $\displaystyle \int 5a \times u \ \mathrm{d}u = 5a \times \frac{1}{2}u^2 = \frac{5}{2}au^2 + k$

where and are constants.

This example is no different.

Thanks.

BobbJo · 08-02-2018 13:13

(Original post by Darke1231)
Attached below an integration problem, giving me a hard time. Could someone solve it step by step and in details.https://cdn.discordapp.com/attachmen...25/unknown.png

(Original post by Darke1231)
As, I mentioned ; I did that already here's what I ended up with.

Hence,that's why am asking for the solution

I'm not sure where you mentioned that.

Darke1231 · 08-02-2018 19:44

(Original post by BobbJo)

I'm not sure where you mentioned that. I must be blind. I feel bad that I can't read minds.

I mentioned it was giving me a hard time, yeah, suck to be you, bud.

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