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Determine $G(x)$.$$G(x)=x^{2} \int_{0}^{x} t^{3} d t$$

$$\frac{3}{2} x^{5}$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 7

Substitution and Properties of Definite Integrals

Integrals

Missouri State University

Harvey Mudd College

Baylor 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.

01:43

Determine $G(x)$.$$G(x…

01:57

Find $G^{\prime}(x)$$$…

01:53

00:51

Find $G^{\prime}(x)$.$…

00:43

01:20

02:15

01:15

Find $G^{\prime}(1),$ wher…

02:05

Please answer ASAP.

0:00

How to solve this question…

03:57

FInd the integral

Okay, so this problem is asking you to do the derivative G. Prime of X. And what's interesting is they define G of X as X squared Times The integral of zero to X of T cubed Gt. The reason why I say this is interesting is there's two things in the problem. First of all, there's a multiplication right here, which is telling me that I better do the product is a product, so do the product rule. And then when I'm asked to do the derivative of the right side, Um this is the fundamental theorem of calculus where you do the derivative and the anti derivative, cancel each other out. So I think I'm ready to do this. That product rule says to take the derivative of the left side, which would be two X. You leave the right side of my own. So that's why I'm leaving it as zero to X of T huge Gt and then minus now leave X squared alone. And when you do the derivative of the right side, it's the fundamental theorem of calculus where the derivative and the anti drip of cancer. So the x replaces the T. Uh T cute. Uh Yeah. Now you can clean this up. I don't know how many teachers would really require you to clean this up. But if you look at this middle piece you could figure out that anti derivative will be 1/4 T to the fourth power from zero to X. And so as you plug in X. Uh A better answer for the this problem would be to X times 1, 4th times X to the 4th And plugging in zero is going to get you zero. So I guess I'll go ahead and write zero. But You know, kind of voice of time minuses. And don't forget you can add your exponents as well. So I'm looking at two times 1/4 as one half X to the fifth minus another X to the fifth. Well 1/2 -1. Give me This should have been a plus room. I think the product rule says to add my mistake, so plus one half plus one would give me three halves X to the fifth. I knew I messed up somewhere. The product rule has to add between the two pieces Derivative of the left side. Leave the right side of 1 plus leave the left side on the right side. But anyway we got there. There's your right answer.

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