Improper Integrals with finite limits

[Silvanoshei]

Was doing a problem and got confused on some evaluation issues of the limits. I got a different answer than my book.

\(\displaystyle \left[\frac{5}{4}(x-2)^{\frac{4}{5}}\right]_{t}^{6}\)

How did that turn into this?

\(\displaystyle \frac{5}{\sqrt[5]{4}}-\frac{5(t-2)^{\frac{4}{5}}}{4}\)

I understand how (4)4/5 power became 5sqrt4, but what happened to the 5/4 in front?

 

[srmichael]

Was doing a problem and got confused on some evaluation issues of the limits. I got a different answer than my book.

\(\displaystyle \left[\frac{5}{4}(x-2)^{\frac{4}{5}}\right]_{t}^{6}\)

How did that turn into this?

\(\displaystyle \frac{5}{\sqrt[5]{4}}-\frac{5(t-2)^{\frac{4}{5}}}{4}\)

I understand how (4)4/5 power became 5sqrt4, but what happened to the 5/4 in front?

The did the following:

\(\displaystyle \dfrac{5}{4}(x-2)^{\frac{4}{5}}\)

Plug in x = 6:

\(\displaystyle \dfrac{5}{4}(6-2)^{\frac{4}{5}}\)

\(\displaystyle \dfrac{5}{4}(4)^{\frac{4}{5}}\)

Use exponent rule of \(\displaystyle \dfrac{a^m}{a^n}=a^{m-n}\)

\(\displaystyle 5(4)^{(\frac{4}{5}-1)}\)

\(\displaystyle 5(4)^{(\frac{4}{5}-\frac{5}{5})}\)

\(\displaystyle 5(4)^{\frac{-1}{5}}\)

\(\displaystyle \frac{5}{\sqrt[5]{4}}\)

 

[Silvanoshei]

The did the following:

\(\displaystyle \dfrac{5}{4}(x-2)^{\frac{4}{5}}\)

Plug in x = 6:

\(\displaystyle \dfrac{5}{4}(6-2)^{\frac{4}{5}}\)

\(\displaystyle \dfrac{5}{4}(4)^{\frac{4}{5}}\)

Use exponent rule of \(\displaystyle \dfrac{a^m}{a^n}=a^{m-n}\)

\(\displaystyle 5(4)^{(\frac{4}{5}-1)}\)

\(\displaystyle 5(4)^{(\frac{4}{5}-\frac{5}{5})}\)

\(\displaystyle 5(4)^{\frac{-1}{5}}\)

\(\displaystyle \frac{5}{\sqrt[5]{4}}\)

Not following, I usually use that rule when doing something like \(\displaystyle \frac{x^{3}}{x^{2}} = x\)
​Where is the 2nd exponent coming from, where is the division???

 

[srmichael]

Not following, I usually use that rule when doing something like \(\displaystyle \frac{x^{3}}{x^{2}} = x\)

Let me spell it out a little clearer:

\(\displaystyle \dfrac{5}{4}(4)^{\frac{4}{5}}\)

\(\displaystyle \dfrac{5 \cdot (4)^{\frac{4}{5}}}{4^1}\)

\(\displaystyle 5 \cdot (4)^{\frac{4}{5}-1}\)

and continue...

 

[Silvanoshei]

Let me spell it out a little clearer:

\(\displaystyle \dfrac{5}{4}(4)^{\frac{4}{5}}\)

\(\displaystyle \dfrac{5 \cdot (4)^{\frac{4}{5}}}{4^1}\)

\(\displaystyle 5 \cdot (4)^{\frac{4}{5}-1}\)

and continue...

Thank you, I'm following the logic now. Last question, -1/5 power to 4 is 5 sqrt 4, but what if it was -2/5 power? Would that be 5 sqrt 8?

 

[DrPhil]

Thank you, I'm following the logic now. Last question, -1/5 power to 4 is 5 sqrt 4, but what if it was -2/5 power? Would that be 5 sqrt 8?

Lets say "5th root" instead of "5 sqrt"

The 5th root of 4 is 4 to the power (1/5): \(\displaystyle 4^{1/5} = \sqrt[5]{4}\)

The power (2/5) can be the square of the 5th root, or it may be written as the 5th root of the square:

\(\displaystyle \displaystyle 4^{2/5} = [4^{2}]^{1/5} = [4^{1/5}]^2\)

......\(\displaystyle = 16^{1/5} = [\sqrt[5]{4}]^2 = \sqrt[5]{16}\)

A negative in the exponent makes all those denominators.