Can some help with question 2 and 4? I will post document.

- anonymous

Can some help with question 2 and 4? I will post document.

- schrodinger

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- Akshay_Budhkar

A] Propose a value for the following
definite integral and give an interpretation
for the result : is nothing but the area under the curve from the values x=6 to x=8

- saifoo.khan

@Akshay_Budhkar , we can find that by simply using area of a square? right?

- Akshay_Budhkar

yes 8^2=64 =D

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## More answers

- saifoo.khan

Perfect. ;D

- saifoo.khan

exactly..

- Akshay_Budhkar

aha the y-axis is degrees / hour! not degrees!

- saifoo.khan

Lol.

- saifoo.khan

It's not easy. haha

- Akshay_Budhkar

u already did :P i am trying

- saifoo.khan

haha.

- anonymous

Amistre64 was helping me on this question but I was still lost so I was hoping if someone else also explained it maybe I would understand.

- Akshay_Budhkar

ok i get it

- Akshay_Budhkar

lets start with A

- Akshay_Budhkar

are you clear with A ?

- saifoo.khan

@Akshay_Budhkar idk but i feel like we have to find some derivatives?

- anonymous

Yes I understand what you said with A.

- Akshay_Budhkar

@saifoo.khan yes

- Akshay_Budhkar

ok now let degrees be d and time be t
so on y axis we have d/t and x axis we have t
Now till x= 1 it is a straight line, with a slope -8 and intercept -8
so the equation is y=mx+c
that is y=-8x-8
that is d/t=-8t-8
d=-8t^2=8t

- Akshay_Budhkar

*d=-8t^2-8

- Akshay_Budhkar

Now for minima or maxima the first derivative of the function is o
therefore\[{d(d) \over dt }=0=-16t-8 \therefore t= 1/2\]

- Akshay_Budhkar

Now to find if it is a minima or a maxima we take the second derivative. if it is negative it is a maxima if positive minima ( mostly i maybe mixing them ) so in this case it is -16 which is negative. therefore it is a maxima at 1/2

- anonymous

Thank you so much I understood that! Would you be able to help me with question 4 by chance?

- Akshay_Budhkar

ok lemme see, is it one unit for each square right?

- anonymous

That is what I understand.

- Akshay_Budhkar

ok lets start with the simplest
C. f(3) = 5 any doubt?

- anonymous

Yes, thats what I see.

- Akshay_Budhkar

ok now A. f'(-2)
Now i believe that at a given point \[f'(x)={dy \over dx} = { y \over x}\] if we estimate. so f'(-2) =-2/-2 = 1

- anonymous

Okay, that makes sense to me.

- Akshay_Budhkar

for D its the same logic, area under the curve. i believe you can assume the graph to be a triangle and calculate the area

- Akshay_Budhkar

is it unclear?

- anonymous

So you think D would be =-2 to 4?

- Akshay_Budhkar

the area under the graph , from -2 to 4

- anonymous

Okay

- Akshay_Budhkar

Can you solve it?
For the other questions i would personally plot a graph for f'(x)

- anonymous

f"(x)=d/dx*f'(x)

- Akshay_Budhkar

yes i know. Now to plot the graph u must know that it is zero at minima and maxima, and u can fidn approximate values by using the approximation of y/x, and plot the graph of f'

- Akshay_Budhkar

then after i get the graph, f"' will be obtained by using y/x again and the last one by area under curve in that graph

- anonymous

Sorry I am just thinking. Okay, Thanks.

- Akshay_Budhkar

yup give it a thought, and lemme know if ustuck =D and u can tag me in some other question if u get some similar doubt =D

- anonymous

Okay, thank you so much for your help!

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