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Determine $G(x)$.$$G(x)=\int_{0}^{x} t^{2} d t$$

$$x^{2}$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 7

Substitution and Properties of Definite Integrals

Integrals

Missouri State University

Campbell University

Oregon State University

University of Nottingham

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.

02:46

Determine $G(x)$.$$G(x…

01:56

01:53

Find $G^{\prime}(x)$$$…

01:57

01:20

00:51

Find $G^{\prime}(x)$.$…

00:43

02:15

01:22

01:51

02:05

Please answer ASAP.

All right. So our task is to find the derivative of this function G prime of X. Um So there's a shortcut, but you might not really understand the shortcut. Eso I'll go ahead and define Thea integral first. So let G of X be defined as this, um the integral from zero x of t squared DT. Well, if you did that, you would add one to the exponents and multiply by the reciprocal from Zero Toe X. And now you can plug in those bounds which you would plug in X for tea on if you plug in 00 cute is still 01 3rd of zero is still zero, and there's no need to write minus zero here. I guess if you wanted to, you could. So that's what G of X is equal to. Well, now, if we do the derivative of that, what you would do is bring that three in front and three times one third will cancel each other out. So you're left with one which you don't have to write one and you subtract one from your exponents. You get X squared on. The derivative of zero is still zero. So this is your answer. But if you follow the fundamental theorem of calculus what you could have done, it is understood that the derivative and the anti driven would just cancel out. And you could just plug in this bound. Now, there is the chain role in here except the derivative of X. Just one. So you don't actually have to worry about that. This answer is just X squared, which was worked out like this. But again, there is a shortcut If your teacher teaches that.

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