Arbitrary Integer in Integral

[eric_f]

Hi All,

I'm working on a problem that states:



I came to an answer of: n ∫ √u du

Can someone let me know if I'm on the right track? I just solved for "dx" during the u-substitution and put du's constant in front of the integral. Is it really that easy or am I borking it up somewhere?


Thanks!

 

[MarkFL]

Using the suggested substitution, we find:

\(\displaystyle u=1+x^{\frac{1}{n}}\,\therefore\,du=\dfrac{1}{n}x^{\frac{1}{n}-1}\,dx=\dfrac{1}{n}x^{\frac{1}{n}}\cdot\dfrac{1}{x}\,dx\)

Now, from the substitution, we find:

\(\displaystyle x^{\frac{1}{n}}=u-1\)

\(\displaystyle x=(u-1)^n\)

Hence:

\(\displaystyle dx=n(u-1)^{n-1}\,du\)

 

[eric_f]

I understand all of this, but...

Using the suggested substitution, we find:

\(\displaystyle u=1+x^{\frac{1}{n}}\,\therefore\,du=\dfrac{1}{n}x^{\frac{1}{n}-1}\,dx=\dfrac{1}{n}x^{\frac{1}{n}}\cdot\dfrac{1}{x}\,dx\)

Now, from the substitution, we find:

\(\displaystyle x^{\frac{1}{n}}=u-1\)

\(\displaystyle x=(u-1)^n\)

Here's where I get lost. Can someone elaborate on the manipulation that is taking place?

Hence:

\(\displaystyle dx=n(u-1)^{n-1}\,du\)

 

[Deleted member 4993]

I understand all of this, but...



Here's where I get lost. Can someone elaborate on the manipulation that is taking place?

That is simple differentiation!!

y = xⁿ

dy/dx = n*x^n-1

dy = n*x^n-1 dx

 

[MarkFL]

We have:

\(\displaystyle du=\dfrac{1}{n}x^{\frac{1}{n}}\cdot\dfrac{1}{x}\,dx\)

or

\(\displaystyle dx=\dfrac{nx}{x^{\frac{1}{n}}}\,du\)

Now, from the substitution, we find:

\(\displaystyle x^{\frac{1}{n}}=u-1\)

\(\displaystyle x=(u-1)^n\)

Hence:

\(\displaystyle dx=n(u-1)^{n-1}\,du\)

 

[eric_f]

I get that it's differentiation, thanks.

I'm probably just making this more difficult than I have to.

So would the new integral be \(\displaystyle n(u-1)^{n-1}\int\sqrt{u}\,\,du\) ?

If not, I'm stuck.

 

[MarkFL]

I get that it's differentiation, thanks.

I'm probably just making this more difficult than I have to.

So would the new integral be \(\displaystyle n(u-1)^{n-1}\int\sqrt{u}\,\,du\) ?

If not, I'm stuck.

You cannot bring a factor involving the variable outside of the integral...you want:

\(\displaystyle \displaystyle n\int \sqrt{u}(u-1)^{n-1}\,du\)

 

[eric_f]

You cannot bring a factor involving the variable outside of the integral...you want:

\(\displaystyle \displaystyle n\int \sqrt{u}(u-1)^{n-1}\,du\)

OK, that makes sense. Thanks for all the help!