Fundamental Theorem of Calculus

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The discussion centers on applying the Fundamental Theorem of Calculus to solve a problem involving definite integrals. Participants clarify that the anti-derivative should not be used for the initial evaluation of F(5), as it leads to confusion regarding the area under the curve. Instead, they emphasize that F(5) represents the integral from 5 to 5, which equals zero, since there is no area between the same vertical lines. For subsequent parts of the problem, the correct approach involves using the theorem to find derivatives rather than direct substitution. Overall, understanding the theorem's application is crucial for solving the problem accurately.
  • #31
So would the answer be:

x^2 + 144?
 
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  • #32
For what? You should be writing an equation; i.e., something with = in it.
 
  • #33
I am completely lost :(
 
  • #34
You're given that ##F(x) = \int_5^x \sqrt{t^2 + 144}~dt##

b) Find F'(5)

To do this part, first find F'(x), and then evaluate this derivative at x = 5. This is where the Fund. Thm. of Calculus comes into play.
 
  • #35
I'm comparing my answers with a similar problem in the book and there aren't any equations, only numerical answers.
 
  • #36
But your work should consist of equations, like this:

F'(x) = <some function of x that you work out>
F'(5) = <some number>

If you just throw something up in a post, I have no idea what you are doing.
 
  • #37
I understand the bit about the Fundamental Theorem of Calculus coming into play but I thought that you just substitute the bounds where the variables are. For example, the book gives us the example of:

Integral with lower bound of 1 and upper bound of x, (t^3)dt. The answer is given as merely x^3
 
  • #38
ralfsk8 said:
I understand the bit about the Fundamental Theorem of Calculus coming into play but I thought that you just substitute the bounds where the variables are. For example, the book gives us the example of:

Integral with lower bound of 1 and upper bound of x, (t^3)dt. The answer is given as merely x^3
OK, let me ask you: What does the answer you wrote represent?
 
  • #39
Mark44 said:
IMHO, memorizing that formula is NOT a good idea, especially if that memorization comes at the expense of understanding.

Point taken I suppose haha, but grant me that I'm not going to write out and explain the whole proof of the theorem lol.
 
  • #40
Or, in other words, what question is x3 the answer to?
 
  • #41
Does it mean that the function is differentiable within those bounds?
 
  • #42
What function? Try to ask questions that are more precise.
 
  • #43
Okay I finally got the answer but only by comparing them to other online resources. I'm still not entirely sure on how to do the actual problem. I wouldn't mind discussing this further but if you guys need to go do other things, that's okay. Thanks for the help anyway.
 
  • #44
ralfsk8 said:
Okay I finally got the answer but only by comparing them to other online resources. I'm still not entirely sure on how to do the actual problem. I wouldn't mind discussing this further but if you guys need to go do other things, that's okay. Thanks for the help anyway.

ralfsk8,

Suppose that G(t) is an anti-derivative of \displaystyle \sqrt{t^2+144}\ .

We write that as \displaystyle G(t)=\int\,\sqrt{t^2+144}\ dt\ .

So if we have a definite integral such as \displaystyle \int_{a}^{b}\,\sqrt{t^2+144}\ dt\,, we can evaluate that as G(b) - G(a), according to the Fundamental Theorem of Calculus. Correct?

Now, in the case of the problem in this thread, we have:
\displaystyle F(x)=\int_{5}^{x}\,\sqrt{t^2+144}\ dt\,=G(x)-G(5)\ .​

Therefore, \displaystyle F&#039;(x)=G\,&#039;(x)\,, since G(5) is a constant.

But G(x) is the anti-derivative of \displaystyle \sqrt{x^2+144}\,, so that \displaystyle G\,&#039;(x)=\sqrt{x^2+144}\,. Correct?

Therefore, \displaystyle F&#039;(x)=\sqrt{x^2+144}\,.
 

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