Question

what is the area between the parabol and the line ? on picture

Answer 1

integrate from 0 to 2 bigger minus smaller

Answer 2

no latex? ---goes into shock and disbelief---

Answer 3

int 0 to 2 -4

Answer 4

so i guess it is
\[\int_0^2 2x-x^2dx\]

Answer 5

well
i thoug ishuld firstly foun s1 the triangle
then i think i can integrate

Answer 6

laptop hard for me to type on. just trying to explain

Answer 7

but dunno the limits :( and what shuld i integrate :(

Answer 8

well i thoug ishuld firstly foun s1 the triangle then i think i can integrate

That's what I would do...

Answer 9

whew! the universe is back on track

Answer 10

what on earth?

Answer 11

thougt like that too estudier :D

Answer 12

estudier is having a stroke

Answer 13

the area of the triangle is 4

Answer 14

Triangle area is 4, why integrate something difficult if u don't need to?

Answer 15

so we just shuld find the total area, s1 + the area between the parabol and the line

Answer 16

call the para medics

Answer 17

in brief; its the area under the parabola minus the area of the triangle ..

Answer 18

yes..

Answer 19

the limits of integration, unless otherwise stated, are prolly where the curves meet

Answer 20

i thought that but i culd not find the integral :d

Answer 21

oooooooooooooooh i see what you mean. yes we get 4 from triangle whew

Answer 22

finaly xd

Answer 23

then integrate
\[\int_0^2 4-x^2dx\]

Answer 24

so to find the area, what shuld the integration be?

Answer 25

hmm
let me do that

Answer 26

o to 2

Answer 27

I think...

Answer 28

Seems like, just calculate intersection...

Answer 29

the limits of integration are where:

4-2x = 4-x^2 ; solve for x

Answer 30

get
\[8-\frac{8}{3}\] subtract 4 get
\[4-\frac{8}{3}\] whatever that is

Answer 31

x = 4 here, and y = 2 here..

Answer 32

ohohoh ... now do it in celtic runes ;)

Answer 33

:) so its 4 to 2? or 0 to 2?

Answer 34

2x ¨= x^2 right?

Answer 35

we could just do it from y = 0 to 4 :)

Answer 36

no need to stick with the x axis

Answer 37

o and 2

Answer 38

i should just shut up and let sensei finish, but this is too much fun. the limits of integration are right in the picture. set the equations equal and solve
you will see that just like in the picture they intersect at x = 0 and x = 2

Answer 39

mhmh

Answer 40

-x^2 +2x = 0 -> x is 0 or 2

Answer 41

lol .... green sensei or blue sensei?

Answer 42

what estudier said, but also wht is in the picture you provided.

Answer 43

so can we focus on questions please?

Answer 44

we are, we are ... but there are a few ways to go about it ...

Answer 45

well just find total area, total area - 4 = S!!

Answer 46

\[\color{green}{\text{green sensei}}\]

Answer 47

So what's 4x- x^3/3 at 2?

Answer 48

a ) 3/2, b) 5/2, c)4/3, d) 7/3, e)9/4

Answer 49

hmm..

Answer 50

ok bigger function is
\[4-x^2\] smaller is
\[2-2x\] subtract to get
\[-x^2+2x=2x-x^2\] then integrate
\[\int_0^2 2x-x^2dx\] as i write up top

Answer 51

solved in very first line. take antideriviative, plug in 2 and be done

Answer 52

so 4r - (r^3 / 3) we have :)

Answer 53

for 2, its 16/3

Answer 54

no it is
\[x^2-\frac{x^2}{3}\]

Answer 55

plug in 2, get
\[4-\frac{8}{3}\]

Answer 56

whops right :D
i wrote worng...!!

Answer 57

.. its 4/3 now what?

Answer 58

now you are done

Answer 59

oO
solution is 4/3??

Answer 60

smaller is 2-2x?

Answer 61

thought it was 4-2x

Answer 62

well m2,buy its 2-r^2..

Answer 63

It is 4-2x so its 8-8/3 -4, right?

Answer 64

God, I hate numbers, give me algebra any day of the week

Answer 65

and one more question, where we got this 2 - x^2 from, i juts don't got it :)

Answer 66

I think we are mixing up methods here.

Answer 67

It's what i said to start with:

Int 4 -x^2, 0 and 2 is 4x -x^3/3 = 8-8/3 then knock off the triangle of 4.

Answer 68

lorda mercy you still working this problem? integral is the very top post. it is not
\[\int 2-x^2dx\] it is
\[\int 2x-x^2dx\]

Answer 69

U said smaller is 2-2x, it isn't, it's 4-2x

Answer 70

Your integral gives same result as mine ie 4-8/3 = 4/3 for the desired area.