Calculating the definite integral using FTC pt 2

[Cjosh]

Homework Statement

Sorry that I am not up on latex yet, but will describe the problem the best I can.
On the interval of a=1 to b= 4 for X. ∫√5/√x.

Homework Equations

The Attempt at a Solution

My text indicates the answer is 2√5. I have taken my anti derivative and plugged in b and subtracted plugged in a. Perhaps I am having an issue with the exponents. Thanks for any advice.

 

[stevendaryl]

The general rule for the integral of powers of $x$ is:

$\int_{a}^{b} x^n dx = \frac{x^{n+1}}{n+1}|_a^b = \frac{b^{n+1}}{n+1} - \frac{a^{n+1}}{n+1}$

What is the value of $n$ for your problem?

 

[Cjosh]

I have ∫5^(1/2) * x^(-1/2). then get [10^(3/2)/3]*[2x^(1/2)]. Not sure what you mean by n value for this problem. My interval a,b is 1,4?

 

[Ray Vickson]

Ca

I have ∫5^(1/2) * x^(-1/2). then get [10^(3/2)/3]*[2x^(1/2)]. Not sure what you mean by n value for this problem. My interval a,b is 1,4?

Can you write $1/\sqrt{x}$ as $x^n$ for some $n$?

 

[stevendaryl]

I have ∫5^(1/2) * x^(-1/2). then get [10^(3/2)/3]*[2x^(1/2)]. Not sure what you mean by n value for this problem. My interval a,b is 1,4?

You already got it: $n=\frac{-1}{2}$.

You seem to be getting confused about the factor $5^\frac{1}{2}$. Why don't you first figure out the answer to the simpler problem:

$\int_1^4 x^{-\frac{1}{2}} dx$

 

[Cjosh]

You already got it: $n=\frac{-1}{2}$.

You seem to be getting confused about the factor $5^\frac{1}{2}$. Why don't you first figure out the answer to the simpler problem:

$\int_1^4 x^{-\frac{1}{2}} dx$

Answering that problem I get the anti-derivative of 2x^(1/2). subtracting a from b I get 3.

 

[Ray Vickson]

Answering that problem I get the anti-derivative of 2x^(1/2). subtracting a from b I get 3.

OK, so now put back the $\sqrt{5}$ in the correct way.

 

[stevendaryl]

Answering that problem I get the anti-derivative of 2x^(1/2). subtracting a from b I get 3.

Well, that's not correct. Could you post your calculation?

 

[Cjosh]

Well, that's not correct. Could you post your calculation?

My bad, I re-did it and got 2. I still get the anti derivative of 2x^(1/2).

 

[stevendaryl]

My bad, I re-did it and got 2. I still get the anti derivative of 2x^(1/2).

Now, multiply by $\sqrt{5}$, and what do you get?