How Does the Fundamental Theorem of Calculus Apply to Derivatives of Integrals?

[Karol]

Homework Statement

Homework Equations

$$F(x)=\int_a^x f(x),~~F'(x)=f(x)$$

The Attempt at a Solution

In F'(x), x is at the end of the domain a-x, so, in my function $~\cos(x^2)~$ i also have to take the end of the domain, and it's 2x, so F'(x)=cos(4x²), but it's not enough.
The answer is F'(x)=2cos(4x²). i have to find F'(x) at x, not at 2x, so i don't know how

 

[Mark44]

Homework Statement

View attachment 198424
View attachment 198425

Homework Equations

$$F(x)=\int_a^x f(x),~~F'(x)=f(x)$$

Your relevant equation isn't quite right. This part of the Fund. Thm. of Calculus is usually presented this way:
$F(x) = \int_a^x f(t)~dt \Rightarrow F'(x) = f(x)$
In short, you have x appearing as both a limit of integration and as the variable of integration. Further, you have omitted dt.

The Attempt at a Solution

In F'(x), x is at the end of the domain a-x, so, in my function $~\cos(x^2)~$ i also have to take the end of the domain, and it's 2x, so F'(x)=cos(4x²), but it's not enough.

The right term is "interval," not domain.

The answer is F'(x)=2cos(4x²). i have to find F'(x) at x, not at 2x, so i don't know how

$F(x) = \int_1^{2x} \cos(t^2)~dt$
Let u = 2x, so du/dx = 2
$F'(x) = \frac d {du} \int_1^u \cos(t^2)~dt \cdot \frac{du}{dx}$
Can you continue with this?

 

[Karol]

In short, you have x appearing as both a limit of integration and as the variable of integration.

I don't understand why to use x and t, and not only x. the graph is in the xy plane, and the domain is on the x axis.
But i couldn't separate between the limit of integration and the variable, as you did in the chain rule, if they were one, so that's the reason? i guess not only.
$$F'(x) = \frac d {du} \int_a^u \cos(t^2)~dt \cdot \frac{du}{dx}=2\cos(t^2)$$

 

[Mark44]

I don't understand why to use x and t, and not only x. the graph is in the xy plane, and the domain is on the x axis.

Because in the Fund. Thm. of Calculus, you have $F(x) = \int_a^x f(t)~dt$, so that F is a function of x, because x shows up as a limit of integration. With a different value of x, you get a different value for F(x). The variable of integration, as I wrote the equation above, and as this theorem almost always appears in textbooks, is t, a so-called dummy variable. Instead of f(t) dt, you could write f(r) dr, f(v) dv, whatever.

For the function in the integrand of your problem, the domain is all real numbers. The interval of integration, which isn't the same as the domain of a function, is the interval [a, x], which could be an interval on the t-axis.

But i couldn't separate between the limit of integration and the variable, as you did in the chain rule, if they were one, so that's the reason? i guess not only.
$$F'(x) = \frac d {du} \int_a^u \cos(t^2)~dt \cdot \frac{du}{dx}=2\cos(t^2)$$

No, not quite. You should end up with $2\cos(u^2)$, but what is u? I wrote it down earlier.

 

[Ray Vickson]

I don't understand why to use x and t, and not only x. the graph is in the xy plane, and the domain is on the x axis.
But i couldn't separate between the limit of integration and the variable, as you did in the chain rule, if they were one, so that's the reason? i guess not only.
$$F'(x) = \frac d {du} \int_a^u \cos(t^2)~dt \cdot \frac{du}{dx}=2\cos(t^2)$$

You cannot have the same x changing and not changing in the same problem at the same time! In the integration $\int f(x) \, dx$, $x$ is changing---go back and review the definition of the integral to see exactly how. In the integration $\int_0^x f(t) \, dt$, $x$ is not changing (but $t$ is).

 

[Karol]

Thank you Mark44 and Ray