Calculating the definite integral using FTC pt 2

In summary: Now, multiply by \sqrt{5}, and what do you get?In summary, the student attempted to solve a homework problem that required them to integrate powers of x, but they were having difficulty with the exponent portion of the equation. After re-doing the problem, they ended up with the correct answer of 2x^(1/2).
  • #1
Cjosh
6
0

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.
 
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  • #2
The general rule for the integral of powers of [itex]x[/itex] is:

[itex]\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}[/itex]

What is the value of [itex]n[/itex] for your problem?
 
  • #3
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?
 
  • #4
Ca
Cjosh said:
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##?
 
  • #5
Cjosh said:
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: [itex]n=\frac{-1}{2}[/itex].

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

[itex]\int_1^4 x^{-\frac{1}{2}} dx[/itex]
 
  • #6
stevendaryl said:
You already got it: [itex]n=\frac{-1}{2}[/itex].

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

[itex]\int_1^4 x^{-\frac{1}{2}} dx[/itex]

Answering that problem I get the anti-derivative of 2x^(1/2). subtracting a from b I get 3.
 
  • #7
Cjosh said:
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.
 
  • #8
Cjosh said:
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?
 
  • #9
stevendaryl said:
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).
 

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  • #10
Cjosh said:
My bad, I re-did it and got 2. I still get the anti derivative of 2x^(1/2).

Now, multiply by [itex]\sqrt{5}[/itex], and what do you get?
 

Related to Calculating the definite integral using FTC pt 2

1. What is the fundamental theorem of calculus part 2?

The fundamental theorem of calculus part 2 states that if a function f is continuous on an interval [a,b] and F is any antiderivative of f on [a,b], then the definite integral of f from a to b is equal to F(b) - F(a).

2. How is the definite integral calculated using FTC pt 2?

The definite integral is calculated using FTC pt 2 by first finding the antiderivative of the given function. Then, plug in the upper and lower limits of integration into the antiderivative and subtract the value at the lower limit from the value at the upper limit.

3. What is the difference between FTC pt 1 and FTC pt 2?

FTC pt 1 deals with the relationship between the derivative and the indefinite integral, while FTC pt 2 deals with the relationship between the definite integral and the antiderivative. In other words, FTC pt 1 is used to find the antiderivative of a function, while FTC pt 2 is used to calculate the definite integral of a function.

4. Can FTC pt 2 be used to calculate the definite integral for any type of function?

No, FTC pt 2 can only be used for continuous functions. If the function is not continuous, then other methods such as Riemann sums or numerical integration techniques must be used to calculate the definite integral.

5. How is the definite integral related to the area under a curve?

The definite integral is related to the area under a curve by representing the area as the sum of infinitely many small rectangles with width dx and height f(x). The definite integral then calculates the total area by taking the limit as dx approaches 0.

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