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VIRGIN9:50 PMCalculusHere f(x) = VxF(r)=Zr Anti derivative of f (x)is From (1) f Vxdr = F(9)-F() =(Gop')-(oy') 52Vxdx Hence;Was this solution helpful? 114...

Question

VIRGIN9:50 PMCalculusHere f(x) = VxF(r)=Zr Anti derivative of f (x)is From (1) f Vxdr = F(9)-F() =(Gop')-(oy') 52Vxdx Hence;Was this solution helpful? 114 51Chapter 5.3Problem 23EStep

VIRGIN 9:50 PM Calculus Here f(x) = Vx F(r)=Zr Anti derivative of f (x)is From (1) f Vxdr = F(9)-F() =(Gop')-(oy') 52 Vxdx Hence; Was this solution helpful? 114 51 Chapter 5.3 Problem 23E Step



Answers

$55-59$ Find the derivative of the function.
$$ g(x)=\int_{2 x}^{3 x} \frac{u^{2}-1}{u^{2}+1} d u $$
$$ \left[\operatorname{Hint} \int_{2 x}^{3 x} f(u) d u=\int_{2 x}^{0} f(u) d u+\int_{0}^{3 x} f(u) d u\right] $$

We have y equals Thean Agrawal from cosign next to Synnex of Natural Log one plus two B Devi. Let's separate this into two intervals. Yeah, Whoops. So this needs to go from co signed zero still of natural log one plus two V and now for the second term were able to swap limits of integration. Now we're ready to go ahead and take the derivative. So we take natural log one plus two V V is going to be evaluated at Sine X, and then we still need to multiply by the derivative of co sign. Excuse me, The derivative of site, which is Casa. For the second term, we have natural log one plus two V where V is evaluated a cosine X, and then we still need to multiply by the derivative of co sign, which is negative sign. So I'll switch this negative out front to a positive, and then we'll apply by a sine x. Okay, then I guess let me include brackets to make it absolutely clear that the cosine and sine or not on the inside of the natural log and this is our answer

In Kriegel, You square negative one upon you. Square positive one. Do you two eggs? 23 eggs. So here we have to find the derivative of the function so we can manipulate the definite intrigue. All intrigued. Two x 20 you Squire. Negative one upon you, Squire. Positive one. Do you? Plus and regal 0 to 3 x use. Credible negative. One upon you, Squire. Positive one. Do you? So here in the first room began interchange the limits. So we put here in Niagara Sign and we get 02 in Kriegel. 0 to 2 eggs, you Squire. 91 upon you, Squire. Positive one. The deal. Mhm close and wriggle. 0 to 3 x, you square. Negative one upon you. Square positive one. Do you? Now we can directly apply the fundamental theorem of calculus and write the dairy video. So here, ***, do the incredible 0 to 2 X You Squire. Pull Swan. So, you Squire, Negative one upon you, Squire plus one. Do you upon d X blows D in real 0 to 3 X, You swear? 91 upon you, Squire. Positive one. Do you of phone DX here? We have to use chain rule and we get negative. Lee and real 0 to 2 eggs. You square 91 a phone, you Squire All the t one deal upon d two x time Lee two x upon dx mhm plus indeed and wriggled 0 to 3 x You square negative one of phone, you Squire plus one deal our phone D t X time d three eggs upon X. Now here we have here we have to apply fundamental theorem of calculus. And you can see that here. After using fundamental theorem of calculus, we get here two eggs square negative one upon two X square. Positive one. Here we get to because we know that the delivery of two eggs here we get to plus and please put here Nguru sign so you can see that here. So please put here. Negative sign. Bless three eggs to the power to negative one upon three eggs. Squire And I bothered you one dying three. Now we simplify this and we get three time. Nine X square. Negative One upon nine X square plus one Negative. Two times four X square, 91 upon four X square, Positive one. So it is a final answer

So here we have f of X is equal to the integral from squared x 22 x of 10 inverse t DT. We can split this up and to to integral those. And now let's go ahead and rewrite this. Using our property that allows us to swap the limits of integration remains his property for the second term. That's how we have a negative now. Okay, now we've rewritten f of X. Let's go ahead and use the fundamental theorem of calculus in order to yeah, when the derivative. So we have tanned, inverse evaluated two X and now we're going to multiply by the derivative of two X, which is just to take 10 adverse of square root X multiplied by the derivative of the square root X, which is one half X to the negative one half. Let's just rewrite this so it's absolutely clear minus one half X to the negative one half tan in verse Squared X That's it

The point problem. We want to draw the graph of F and use it to find a rough sketch of the derivative or the anti derivative. So it's going to be F. Of X equals The root of X. to the 4th and it's two X squared plus two And then -1. Yeah. So let's consider the fact that the anti derivative tells us we want this to be the derivative. So if this is going to be our derivative we see that that we have the whole time. The value is positive. However there are gunnery parts of -1 and one where the graph studies off, so we see the graphs can be increasing. It's going to study off at -1, have a slope of zero and then increase again. It's going to reach um this inflection point and then study off again as well once it gets to one and then it's going to keep increasing.


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