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Sketch the area represented by $ g(x) $. Then find $ g'(x) $ in two ways: (a) by using Part 1 of the Fundamental Theorem and (b) by evaluating the integral using Part 2 and then differentiating.

$ \displaystyle g(x) = \int^x_1 t^2\,dt $

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01:05

Frank Lin

06:44

Chris Trentman

Calculus 1 / AB

Chapter 5

Integrals

Section 3

The Fundamental Theorem of Calculus

Integration

Campbell University

Oregon State University

Harvey Mudd College

Baylor University

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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Sketch the area represente…

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$5-6$ Sketch the area repr…

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5-6 Sketch the area repres…

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Find $g^{\prime}(x)$ using…

All right, let's take a look at this function G of X. G of X is equal to the definite integral from T equals 12. T equals X. Of the function T squared D two. What does this mean? All right, first of all, this is going to be our T axis since we have a function of T. Okay, X is just a number on the T axis. It could be 235 negative one. Um So X is just a number but are variable is T. Um Okay, what does the function T squared look like? T squared, looks like this And when T is one, T squared is one. Okay, so this is a graph of T square. No, X is some number To the right of one on the T axis. X is just a number. Now. The indefinite integral between one and X of t squared. That means we are finding the area underneath the t squared function between one An ex. Okay, so the definite integral between one and some number X of t squared DT is the area underneath the t squared function Between the numbers one and X. So, this area in here, if I kind of shaded in. So the area underneath the t squared function between one and some number X is uh the value of this definite integral. Now G is a function of X. So, what this is really meaning is G of X means the area underneath the t squared curve between one and some number X. So G F two means the area underneath the t squared function between one and two. G of three means to area uh underneath the t square curve between one and three. So G is a function of X O k G of X. The integral from one X of t squared DT is really the area underneath the t squared curve between the numbers one and the number X. Now we G fx is a function and we can differentiate, we could find G prime of X. Using the fundamental theorem of calculus, if G fx equals the integral between one and x T square D T then G prime of X is simply the T function T squared evaluated at X. So G prime of X is simply to t squared function evaluated for T equal X. That's a fancy way of saying uh simply plug the X in 40 square instead of t squared X squared. So if Kiev x is equal to the integral between one and X of t square D T. The fundamental theorem of calculus says G prime of X will be instead of t squared X squared. No, the other way uh they want us to find G prime of X is to actually calculate uh this uh definite integral. Sit down here, we're actually going to calculate the definite integral. So G fx in case we're back to G fx GFX is equal to the integral between one and X of t squared DT. So if you want to calculate a definite integral uh we need to find an anti derivative of T square Okay, G of X is equal to the integral definite integral from one X of t square D T. Well the integral T squared is T to the third over three. And since this isn't definite, since this is a definite integral, we have to evaluate to t cubed over three uh between T equals X And T. equals one. That simply means plug in X into this expression and then subtract one, plugged into this expression. So t cubed over three when T is X is executed over three and then subtract uh and then plugging in one for t t cubed over three. When T is one is one cubed over three or just 1/3. Mhm. So that's our G fx function. G f x Okay, is the integral from one X of t squared DT But when we actually do that definite integral we find out that G of X really is executed over three minus one third. Well now we could find G prime of X simply by taking derivative the derivative of this function since G fx equals execute over three minus one third. G prime of X will be the derivative of X cubed over three minus one third. While the derivative of X cubed is three X squared. So the derivative of X cubed over three is three X squared over three minus and derivative of a constant. Like one third is zero, so three X squared over three. The threes cancel and we get just X squared. So take a look at what just happened. We found out that G prime of X is equal to X square. Well, we already knew cheap prime of X was equal to X squared using the fundamental theorem, A calculus. But now actually, you know, Kathleen to definite integral, finding out the actual function for G fx and then taking the derivative of it. We just found a second way of finding G prime of X and of course G prime of X is still G prime of X. So we would expect it to equal X squared both times. Even though we found in two different ways is still the G same G prime of X. So it better still equal the same function which it does. X squared G prime of X equals X squared

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