Fundamental Theorem of Calculus

**mei** · June 23rd 2009, 11:42 AM

Let $g \in C^1$ such that g(0)=0 and g'(0)=1

$\phi(x)=\int g(t).dt$

upper limit: $log x$
lower limit: $x^2-1$

I have to calculate $\phi'$ and $\phi''$ and I have been reading about the fundamental theorem of calculus but I'm not too sure of how I can apply it to calculate $\phi'$ and $\phi''$ . Every example I look at has some expression instead of just g(t) like I have.

(I apologize for not knowing how to put the limits in the integral.)

Thanks in advance.

**skeeter** · June 23rd 2009, 02:02 PM

Originally Posted by mei

Let $g \in C^1$ such that g(0)=0 and g'(0)=1

$\phi(x)=\int g(t).dt$

upper limit: $log x$
lower limit: $x^2-1$

I have to calculate $\phi'$ and $\phi''$ and I have been reading about the fundamental theorem of calculus but I'm not too sure of how I can apply it to calculate $\phi'$ and $\phi''$ . Every example I look at has some expression instead of just g(t) like I have.

(I apologize for not knowing how to put the limits in the integral.)

Thanks in advance.

$\phi(x) = \int_{x^2-1}^{\log{x}} g(t) \, dt$

$\phi'(x) = g(\log{x}) \cdot \frac{1}{x} - g(x^2-1) \cdot 2x$

**Random Variable** · June 23rd 2009, 02:26 PM

Originally Posted by skeeter

$\phi(x) = \int_{x^2-1}^{\log{x}} g(t) \, dt$

$\phi'(x) = g(\log{x}) \cdot \frac{1}{x} - g(x^2-1) \cdot 2x$

But how do you incorporate the given conditions?

**skeeter** · June 23rd 2009, 02:53 PM

Originally Posted by Random Variable

But how do you incorporate the given conditions?

good question ... you think the OP has posted all the info regarding the problem?

**Random Variable** · June 23rd 2009, 02:55 PM

Originally Posted by skeeter

good question ... you think the OP has posted all the info regarding the problem?

I'm leaning towards "no."

**mei** · June 24th 2009, 12:41 AM

Originally Posted by Random Variable

I'm leaning towards "no."

I have posted all the info that was given me.

**HallsofIvy** · June 24th 2009, 01:33 AM

Originally Posted by mei

Let $g \in C^1$ such that g(0)=0 and g'(0)=1

$\phi(x)=\int g(t).dt$

upper limit: $log x$
lower limit: $x^2-1$

I have to calculate $\phi'$ and $\phi''$ and I have been reading about the fundamental theorem of calculus but I'm not too sure of how I can apply it to calculate $\phi'$ and $\phi''$ . Every example I look at has some expression instead of just g(t) like I have.

(I apologize for not knowing how to put the limits in the integral.)

Thanks in advance.

The "fundamental theorem of calculus" says that if $F(x)= \int_a^x f(t)dt$ then F'(x)= f(x).

Here you have an integral but the lower limit is a function of x, not a constant, and the upper limit is a function of x, not x. We can handle the first problem by separating this into two integrals:
$\phi(x)= \int_{x^2-1}^a g(t)dt+ \int_a^{log x} g(t)dt$
where "a" is any specific number. Then
$\phi(x)= \int_a^{log x} (t)dt- \int_a^{x^2-1}g(t)dt$
since swapping the upper and lower limits of integration multiplies the integral by -1.

We take care of the problem with the upper limit by introducting two new variables: let $u(x)= x^2- 1$ and $v= log(x)$ so
$\int_a^{x^2-1} g(t)dt= \int_a^u g(t)dt= f(u)$ , $\int_a^{log(x)}= \int_a^v g(t)dt= h(v)$ , and $\phi(x)= f(u)- h(v)$ .

Now, use the chain rule to find $\frac{d\phi}{dx}$ and $\frac{d^2\phi}{d\phi^2}$ .
$\frac{d\phi}{dx}= \frac{df}{du}\frac{du}{dx}- \frac{dh}{dv}\frac{dv}{dx}$ . Of course, $\frac{df}{du}= g(u)$ and $\frac{dh}{dv}= g(v)$ .

You "put the limits in the integral" by using _{ } for the lower integral and ^{ } for the upper limit. In fact, you can see the code for any LaTex on this forum by clicking on it.

Click on $\int_{x^2-1}^{log(x)} g(t)dt$ and see what you get.

**mei** · June 24th 2009, 02:43 AM

Thank you so much!
I think I was having a bit of a problem on how I was supposed to incorporate the chain rule onto that, but now I get it!

And also thank you for explaining the "nice" way of writing integrals in this forum.

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