Fundamental Theorem of Calculus

[Qube]

Homework Statement

The derivative of an integral with a constant as its lower bound and a function as its upper bound is the function at its upper bound multiplied by the derivative of the upper bound.

The Attempt at a Solution

How come the constant term has no bearing? I understand the derivative is un-doing the integral, but how come the constant term disappears?

Is this why?

https://scontent-b-mia.xx.fbcdn.net/hphotos-prn2/v/1459734_10201135696496253_598181064_n.jpg?oh=a9ce27391aab214e4545d0bddf547aa5&oe=5292F89D

 

[haruspex]

Having x as the variable of integration and in the range is a 'pun'. I.e. they're different variables really, so to make it clearer let's give them different names: $\frac d{dx}\int_{t=1}^{x^2}t.dt$
Increasing x extends the integral at the right-hand end, but doesn't change the left-hand end. So the change in value of the integral does not depend on where the left-hand end is. Draw a picture.

 

[Qube]

Having x as the variable of integration and in the range is a 'pun'. I.e. they're different variables really, so to make it clearer let's give them different names: $\frac d{dx}\int_{t=1}^{x^2}t.dt$

Yes, it's the dummy variable of integration.

Increasing x extends the integral at the right-hand end, but doesn't change the left-hand end. So the change in value of the integral does not depend on where the left-hand end is. Draw a picture.

I love geometric approaches to calculus. Indeed, what you say is true.

 

[Ray Vickson]

Homework Statement

The derivative of an integral with a constant as its lower bound and a function as its upper bound is the function at its upper bound multiplied by the derivative of the upper bound.

The Attempt at a Solution

How come the constant term has no bearing? I understand the derivative is un-doing the integral, but how come the constant term disappears?

Is this why?

https://scontent-b-mia.xx.fbcdn.net/hphotos-prn2/v/1459734_10201135696496253_598181064_n.jpg?oh=a9ce27391aab214e4545d0bddf547aa5&oe=5292F89D

Besides what others have told you, you can apply the general formula:
[tex] \frac{d}{dx} \int_{a(x)}^{b(x)} f(t) \, dt = f(t)|_{t=b(x)} \frac{db(x)}{dx} - f(t)|_{t=a(x)} \frac{d a(x)}{dx}

 

[Ray Vickson]

Homework Statement

The derivative of an integral with a constant as its lower bound and a function as its upper bound is the function at its upper bound multiplied by the derivative of the upper bound.

The Attempt at a Solution

How come the constant term has no bearing? I understand the derivative is un-doing the integral, but how come the constant term disappears?

Is this why?

https://scontent-b-mia.xx.fbcdn.net/hphotos-prn2/v/1459734_10201135696496253_598181064_n.jpg?oh=a9ce27391aab214e4545d0bddf547aa5&oe=5292F89D

Besides what others have told you, you can apply the general formula:
$$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) \, dt = f(t)|_{t=b(x)} \frac{db(x)}{dx} - f(t)|_{t=a(x)} \frac{d a(x)}{dx}$$
For a constant lower limit we have da/dx = 0.