5

Hint F irSt take logS 3_dw (w _ 1)24dx Vj2 6w + 85_dx f exV1 L e-22...

Question

Hint F irSt take logS 3_dw (w _ 1)24dx Vj2 6w + 85_dx f exV1 L e-22

Hint F irSt take logS 3_ dw (w _ 1)2 4 dx Vj2 6w + 8 5_ dx f exV1 L e-22



Answers

Calculate $F(4)$ giventhat $F(1)=3$ and $F^{\prime}(x)=x^{2} .$ Hint: Express
$F(4)-F(1)$ as a definite integral.

All right, So we are being asked what four is knowing what f of one is and knowing that are derivatives of F is X squared. So what that means is they want me to set up an integral that looks like problem is, there could possibly be a constant that I don't know about when I go to do the derivative. But this is what I've got, right? So if I think about this x squared, going backwards for sure is Q 1/3 excuse. That's what I got. But when I check that 1/3 have one to the power of three. Right now, I'm sitting at 1/3. But my answer has to be three. That means I have a constant of some sort, so that when I do, the derivative of that just cancels out anyway. So I have to kind of work backwards here to figure out what my constant Woz be able to figure out what my F four. So what am I missing on 1/3 they get to three. And the answer for that cause I just have to subtract 1/3 from three. So my final derivative, you could use it as two and 2/3 or could use it as 8/3 is my con. So once we know this important piece of information, finding f F four is really easy, because that means my derivative equation is 1/3 the power of three less 8 30 So just pull a gun. And four, we end up with 1/3 four cube plus 8 30 So four Cube is 64 so end up with 64 over three. Unless I eat three that works out to 72 over three, which does bye 2 24 ff four would be the value 24.

Okay. Quite a cool little question here we're were asked to calculate f of four, given the F of one is equal to three and F prime of X is equal to x square. So, I'm going to use the hint to set up a definite integral. Okay. And we know that if we do F of four minus F of one, that will be the integral from one to full the a prime of X. Dx. Ok, because we know that this function over here on the left, big F just has to be the anti derivative of what's in here. Okay, this allows us to say the F of four is equal to we can drop our number in for F of one, which is three. And then on this side we have an integral from 1-4 of x squared dx We've got an integral. We can compute on this side. Okay, this will be X cubed over three From 1 to 4 for for ministry we can apply our fundamental therm of calculus, so four cubed is 60 for over three minus one, cube is 1/3. Okay, that'll give us 63/3, which I believe is 21. So we end up with f of 4 -3 is 21. We can move the three over As if by Magic, we've solved for F 4. 4 must be 24.

Given that we have the whole area of the integral from 2 to 8, and that is 7.3. Let's go ahead and just sketch that out here. I just drew an arbitrary graph, so this is not exactly what it looks like. Of course I'm just using this as an example. But that tells us that the area here under the curve would all be 7.3. And then we're told from 2 to 4 that the area is 5.9. So then this area here would be 5.9. The question is, what is the area than from 4 to 8. So this area here, what would that evaluate too? And to do that, we could simply subtract the total area 7.3 minus 7.3 and then minus our 5.9. And that will end up giving us just go ahead and do that out. Six. That was before and 1.4. Therefore, the missing area here must be 1.4. Because if we were to add that area 1.4 plus 5.9, that would then get us back to 7.3 as well. So that is what the missing area for the integral from 4 to 8 would have to be here.

Okay, So off the property of definite into all states that get me in to change the interval off integral in its value becomes negative. So this means that the integral from A to B off F off X is equal to negative terrace Ian to grow from B to a off off X So when you have the integral from 4 to 5 off f off X t x, this is equal to negative times integral from 5 to 4 off off X dx, which is equal to 36. So therefore of senior 3.6. So therefore, since this is equal to 3.6, this is that equal to negative 3.6. So if we add both the integral Sze wee enough with the integral from 1 to 5 off f off X t X plus the integral from 5 to 4 off f of X d. X, which is equal to 12 minus 3.6, which is equal to eight 0.4


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