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Problem 6 Easy Difficulty

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_0 (2 + \sin t)\,dt $

Answer

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

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Video Transcript

it's good. Were given a function G. And the rest to sketch the area represented by dysfunction and then define the derivative G. Prime in two different ways. The function is G. F. x equals the integral from 0 to X. Of two plus the sign of T. T. T. I had it all over my problem like. Yeah. So the sketch the area represented by G. 1st sketch the function to plus sign T. This is just a vertical shift of the function. Sign T. Up two units. Yeah. And so a sketch of this looks something like this. We start out at the point. Yes, 02. We move up to the point Pi over 2, 3 and back down to the point pi to end the 0.0.3 pi over two -1. Sorry, just one and then two pi two and it repeats itself. Already did all the homework so we can get to the cylinder car. Yeah. Yeah. Uh huh. Yes. And for our shaded area we start out At X equals zero. And we move over to X. For example, if X is somewhere greater than zero, say X is here. I don't know. He should then G represents the area here in green. Uh huh. It's not all doesn't dry like that. Come come evac evaporate. It's pretty soon. No calm. Makes these crusty. If you wouldn't you didn't think. Okay now we want to find G. Print of X in part A. By using part one of the fundamental theorem of calculus over so function a little F of T. Is two plus the sign of T. Nobody really. You haven't seen coming now we'll let big F. F. T. B. The anti derivative. Did you spell guys? She are the function little Fft. Yes. By the first part of the fundamental theorem of calculus, if G. Of X is the integral From 0 to X of two plus the sign of T. Then it follows that the derivative G prime exists and is given by this is the derivative of this integral which is the same as just little F of X. And plugging an F. Into expression little F. Sorry X. Into the expression Little left. This is two plus the sine of X. Oh yeah, he's written, I don't think it's been a stuff really just slinks around like this. So that's part A. Now, as for part B, that's a big it is. I mean like the term big is overused. Okay, I'm not shocked. Yeah. Now, using the second part of the fundamental theorem of calculus, we have that we can also represent G. Fx as the anti derivative big F. Of T evaluated from zero to X. And so, plugging in this is equal to F of X- Big F of zero. In differentiating we get the G prime of X is the derivative of this, which is F prime of X. And then FF zero is a constant. So it's derivative is simply zero. And by definition Big F. Is the anti derivative of a little left, so this is equal to little F of X. Which plugging in. This is two plus the sine of X. So who's who's saying now? That's one way to find it using the second part. So many. Right. Right. That's crazy. They didn't do that, Yeah.