Question

Calculus Question!! Medal!!

Answer 1

Is this familiar to anyone?

Answer 2

it looks like a hyperbola to me

which is in the form

\[\Large \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2} = 1\]

Answer 3

so wouldn't that be option C

Answer 4

yes

Answer 5

Thank you for your input :)

Answer 6

no problem

Answer 7

do you think you could help me with another question?

Answer 8

sure

Answer 9

thank you :) let me post it

Answer 10

@jim_thompson5910

Answer 11

one moment

Answer 12

ok the first step is to plot all the points

see attached

Answer 13

let me know when you're ready for the next step

Answer 14

I'm ready

Answer 15

ok now form rectangles such that the left bottom tip touches the points A through D

like you see in the attached image

Answer 16

point E is left out because it's at the right end of the interval

Answer 17

Find the area of each rectangle. Then add up the areas. What result do you get?

Answer 18

5(-6)+2(-11)+3(-18)+1(-21) like this?

Answer 19

5(-6)+2(-11)+3(-18)+1(-21)=-127

Answer 20

very good

Answer 21

so the approx area is -127

divide this approx area by b-a = 15-4 = 11

Answer 22

-11.545

Answer 23

I'm basing this off of this formula


Average Value of Function f(x) on interval [a,b]
\[\Large V = \frac{1}{b-a}\int_{a}^{b}f(x)dx\]

Answer 24

-11.545 is the final answer

Answer 25

Thank you soo much. I don't want to bother you but it's late and i need help with on more question. Could you help?

Answer 26

yeah

Answer 27

Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x^3 and y = x.

Answer 28

how far did you get?

Answer 29

This is the graph

Answer 30

which region are we looking for? paint in the region with a different color

Answer 31

The regions in orange??

Answer 32

correct

Answer 33

notice how the two pieces are symmetric, so we can find one half and double it to get the total area

Answer 34

sound good so far

Answer 35

notice how the straight line y = x is above the curve y = x^3 on the interval 0 < x < 1

the area of that region is

\[\Large \int_{0}^{1}\left[x - x^3\right]dx\]

agreed? or no?

Answer 36

agreed

Answer 37

so you'll graph `y = x-x^3` then set up 4 rectangles like so

notice how the midpoints A,B,C,D are set up and are used to determine the heights of each rectangle

eg: the x coord of point A is at the midpoint of x = 0 and x = 0.25, the x coord of point B is at the midpoint of x = 0.25 and x = 0.50, etc

Answer 38

find the area of each rectangle. Then add up the 4 areas. Finally double the result to get the approx area of both regions combined

Answer 39

Is this good so far?
0.25(0) + 0.25(0.25)+ 0.25(0.5) + 0.25(0.75)

Answer 40

it's too big

Answer 41

look at my attached image. Look specifically at the y coordinates of A,B,C,D. They determine how high each rectangle is

Answer 42

wait I have to use the y coordinates of A,B,C,D? so like 0.25(0.25) as the first one?

Answer 43

ohhh wait you mean like 0.00195(0.125)

Answer 44

where are you getting 0.00195?

Answer 45

the distance between the x and y coordinates of point A

Answer 46

each rectangle is of width 0.25

the height is the only thing that changes

the height of the first rectangle is the y coord of point A, so the height of the first rectangle is 0.12305

Answer 47

ssoo the fist one is 0.25(0.12305)??

Answer 48

yes

Answer 49

oohh okay let me do the rest

Answer 50

0.25(0.12305) + 0.25(0.32227) + 0.25(0.38086) + (0.25(0.20508)

Answer 51

0.25(0.12305) + 0.25(0.32227) + 0.25(0.38086) + 0.25(0.20508) = 0.257815

geogebra is saying 0.2578125

so there's a bit of roundoff error

I updated the drawing with more decimal places shown

Answer 52

yeah i got .257815

Answer 53

I guess you'll probably round to 2 or 3 decimal points, so it won't matter at the end

Anyways, double the result to get the final answer

Answer 54

0.51563

Answer 55

btw if you're curious, this is how it would look like if you didn't find the area under the curve of y = x-x^3

this method takes a lot more work though

Answer 56

I'm getting 0.51563 also

Answer 57

woah glad I didn't get that question haha. Thanks again for all the help your awesome!

Answer 58

no problem