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Ht(xl= 2 hlx)= O nlx)- ( = 2 1 V 3x2 3xcos(x) Theorem 1{ 03 of Calculus find Ine detivacive nA DWe 11 cos(t) at...

Question

Ht(xl= 2 hlx)= O nlx)- ( = 2 1 V 3x2 3xcos(x) Theorem 1{ 03 of Calculus find Ine detivacive nA DWe 11 cos(t) at

ht(xl= 2 hlx)= O nlx)- ( = 2 1 V 3x2 3xcos(x) Theorem 1{ 03 of Calculus find Ine detivacive nA DWe 1 1 cos(t) at



Answers

Find each derivative using Part $I$ of the Fundamental Theorem of Calculus. $\frac{d}{d t}\left[\int_{0}^{t}\left(3+x^{2}\right)^{3 / 2} d x\right]$

Mhm. To use Part two of the fundamental theorem of calculus, we first need to find the anti derivative and then evaluate the upper and lower limit and find the difference. In this case. It might be easier to find the anti derivative by changing the form of the integrated, the anti derivative is now X. To the negative second over negative two, which will be evaluated from 1 to 3. To simplify this form, we're gonna change it to the opposite of two X squared. For our Integrated that we will still be evaluated from 1 to 3 as we evaluate our into girl. Take care to plug in in the correct order and simplify as you go. In this case negative 1 18 plus one half will simplify to four nights.

So about this question we need to uh find this different degrees using the fundamental theorem of calculus. So that's 0 to 7. Or to integration of two X plus one times two X plus one times minus 1/3 D. X. Uh So to integrate this we can make a simple substitution of two X plus one as let's call it T. So this means that two D. X. Is equal to D. T. Which means that de excess D. T. Or two. Uh So also we have to change the limits as well affects zero, affects zero. This means that the value of T. S two times zero plus one which is nothing but one. And if X is seven or two then T will become two times seven or two plus one which is eight. So the limit changes from The lower limit becomes one because it is no 1-8 from 1 to 8 and two X plus one is nothing more. So it's T raise two minus 1/3 and dX is nothing more DT over to. So this becomes equal to one or two. Is outside here is two minus one or three is uh integration stages to two or three over to over three within the limits of 1 to 8. Uh So this becomes uh this becomes three comes on top, so we have three or four because two times two is four and we have upper limit is eight raised to two or three minus 11 raced towards there is nothing but one and eight cube root is nothing but two and two squares four. So we have four minus one over here And 4 -1 is nothing about three. So we have three times three which is nine, so we have nine or four over here, which is the final answer. Thank you.

The fundamental theorem of calculus two 100 of this. First thing will it's actually going to get this into its proper form by flipping the intervals and to flip the other. Well we have to multiply by negative one to the outside there. So then that's we have here. So now taking the derivative of this, why prime equals It's going to be a negative one times one minus three X. To the third power cooper one plus one minus three acts the second, then it's times the derivative of 1 -3 x. So then the derivative is since negative on since one minus Next to the 3rd, one plus 1 -1 Reacts to the 2nd Period of 1 -3 X is -3. Our final answers three times 1 -3 x. Third over one plus one minus three X squared. That's the interpretive there.

In this question we have to find the integration of the function nowadays. 0-1 in the limits. And the function is Access to the power four x 3 minus two. Into access to the power one by three. Okay now we have to integrate each of them. I integrate access to the bar four by three. So I get access to the power four by three plus one upon four by three plus one further minus two into X. Raised to the power one by three plus one upon one by three plus one. Okay these are the values of integration and we have to put the limits here. So first of all we simplify the value that we don't get is three x 7. Access to deep out seven x 3 minus story in Do three by food express to the power four x 3. Okay and now we have to put the limits here. Okay So I'm going to put here one so I get here three x 7 -2 and 2. 3 x four when I put one Okay and minus when I bought zero you can see that X is replaced by zero. Both of the terms are now in 20. Okay now we have to simplify this. That is three x 7 and -3 x two. Here we can the right details 14 that is the least common multiple of seven and two. And then we got here six minus 21 6 minus 21 is minus 15 so minus 15 divided by 14 is the final answer of this question. Okay thank you.


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