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# Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.$\displaystyle g(x) = \int^x_0 \sqrt{t + t^3} \,dt$

## $g(x)=\sqrt{x+x^{3}}$

Integrals

Integration

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### Video Transcript

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

University of California, Berkeley

Integrals

Integration

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