Thread: What does this calculus question mean?

1. What does this calculus question mean?

Revising for an exam here and looking at last years papers I can across this question -

I am unsure of what actually is 'f' in this case.
Is it (t^2 - 4) or do is f the integral of (t^2 - 4)?

And if it is the integral of (t^2 - 4) how am I supposed to integrate that when the values given are 0 and X. I have never seen an X in this location of the integration notation before so I'm not sure what to do with it?

2. Originally Posted by nukenuts
Revising for an exam here and looking at last years papers I can across this question -

I am unsure of what actually is 'f' in this case.
how am I supposed to integrate that when the values given are 0 and X. I have never seen an X in this location of the integration notation before so I'm not sure what to do with it?
There is not need to do that.
If $g$ is an integrable function then
if $f(x) = \int_0^x {g(t)dt}$ the derivative is simply
$f'(x)=g(x)$.

3. Originally Posted by nukenuts
Revising for an exam here and looking at last years papers I can across this question -

I am unsure of what actually is 'f' in this case.
Is it (t^2 - 4) or do is f the integral of (t^2 - 4)?

And if it is the integral of (t^2 - 4) how am I supposed to integrate that when the values given are 0 and X. I have never seen an X in this location of the integration notation before so I'm not sure what to do with it?
Plato's method is the best way to go about doing this problem, but the "brute force" approach isn't that difficult. Just integrate f(x) and put in the limits in. It's no different from the integral that has an upper limit of, say, 4.
$\displaystyle f(x) = \int_0^x(t^2 - 4)dt = \left . \left ( \frac{1}{3}t^3 - t \right ) \right |_0^x$

Just put the 0 and x in as if they were numbers. (Then you could take your derivative and find out that you took the long way about to get Plato's answer.)

-Dan

4. I am not sure I am picking you up right. I dont see why you are giving me the derivative when the question starts off by giving me an integral
$f(x) = \int_0^x {(t^2 - 4)dt}$

Can you tell me what is actually my 'f' and then how to find the critical points of this f? (I know you use f'(0) to get the critical points I just want to make sure Im not mixing up what you are telling me.)

5. $\displaystyle f(x) = \int_0^x\left(t^2 - 4\right) \, dt$ means $\displaystyle f$ is the integral of $\displaystyle t^2 - 4$ (from $\displaystyle t = 0$ to $\displaystyle t = x$).

And you don't "use f'(0)" to find the critical points of $\displaystle f$, you solve the equation $\displaystyle f ^\prime (x) = 0$ in terms of $\displaystyle x$, which is why Plato and topsquark showed you how to find the derivative of $\displaystyle f(x)$ instead of focusing on evaluating the integral. In other words, don't worry about evaluating the integral as the questions are more concerned with $\displaystyle f ^\prime (x)$ and $\displaystyle f ^{\prime \prime} (x)$.

For example, consider $\displaystyle g(x) = (x + 1)^2$. If asked to find the derivative of $\displaystyle g(x)$, you could expand $\displaystyle g(x)$ like so

$\displaystyle g(x) = x^2 + 2x + 1$

and then find $\displaystyle g ^\prime (x)$ by differentiating each term like so

$\displaystyle g ^\prime (x) = 2x + 2$

or you could simply differentiate $\displaystyle g(x)$ as it is like so

$g ^\prime (x) = 2(x + 1)$

The first method involves an unnecessary expansion of $\displaystyle g(x)$. Similarly, evaluating the integral in your problem is unnecessary because you are more concerned with $\displaystyle f ^\prime (x)$. (Of course, both methods are valid as topsquark showed.)

6. Look here at the subsection titled "Formal Statements."

I am rather surprised to find that you (apparently) don't have the background to do this? This theorem would have been in your reading in the section before any practice problems....

-Dan

7. Originally Posted by topsquark
Plato's method is the best way to go about doing this problem, but the "brute force" approach isn't that difficult. Just integrate f(x) and put in the limits in. It's no different from the integral that has an upper limit of, say, 4.
$\displaystyle f(x) = \int_0^x(t^2 - 4)dt = \left . \left ( \frac{1}{3}t^3 - t \right ) \right |_0^x$

Just put the 0 and x in as if they were numbers. (Then you could take your derivative and find out that you took the long way about to get Plato's answer.)

-Dan
OK, shouldn't that be $( \frac{1}{3}t^3 - 4t )?$

I think I get it now. So if I subbed in the x and the 0 and subracted g(x) - g(0)

I would get $( \frac{1}{3}x^3 - 4x )$

And this is my f function?

And I already have f' as it is given to me at the start as (t^2 - 4)? And basically I dont even need to do the integration as I have already been given the derivative in the question, I just have to replace t with x, so I would use (x^2 - 4) = 0 to get the critical points? And that is basically what Plato told me from the beginning?

Can someone just confirm that I have got all that right? Cheers.

8. Yes, you are correct. It's called the Fundamental Theorem of Calculus. (I believe topsquark provided a link to a Wikipedia article on the topic.)

9. Ok, thanks for the help lads, I understand it now.