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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 $

a. $g^{\prime}(x)=f(x)=x^{2}$

b. $g^{\prime}(x)=x^{2}$

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Missouri State University

Baylor University

University of Michigan - Ann Arbor

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