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Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

$ \displaystyle g(x) = \int^x_0 \sqrt{t + t^3} \,dt $

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00:41

Frank Lin

00:20

Amrita Bhasin

Calculus 1 / AB

Chapter 5

Integrals

Section 3

The Fundamental Theorem of Calculus

Integration

University of Michigan - Ann Arbor

University of Nottingham

Idaho State University

Boston College

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

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Use Part 1 of the Fundamen…

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01:10

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07:16

Use the Fundamental Theore…

01:01

All right, let's go ahead and do this problem. So we are given an integral and we're supposed to find the derivative off this expression with respect to X, using part one off the fundamental theme of calculus. So to do a really quick recap, it basically boils down to the fact that taking the derivative often integral basically undo each other. So if you have an expression that looks like this, what I want you to see that the variable use for the integration part is T But you're plugging in an X. This makes it into a function of X, as you can see over here. So it's just that the relationship between large F and small f right here is that small f is a derivative of large F. Or you could also say that the anti derivative or the integral off small f is large f. Okay, so if you take the derivative off, um, large ffx, it just ends up being small ffx. So from the point of view from here, what it looks like, it just seems like the inte grand in the DT just disappeared. And instead of a t, now you have an ex. Okay, so this is the most straightforward way that you would do this problem so you would have to still deal with situations. Situations such as the product. Will the Kocian rule the change rule because you're still taking a derivative. But when the input is simply X, you don't have to worry about that portion, so you can just think of it as canceling out the integration. So let's apply to this problem if you take the derivative with respect to X off the integral from zero to eggs off a function that takes this form T plus tty cubed inside the square root where the variable is t. Okay, we know that the integration and the derivatives are going thio undo each other. And let's remember that the X is three input that makes this into a function of X. So the result just ends up being X plus X cubed inside the square root. And that's all you need to do

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