Question

Calculate the area of the region bounded by the graphs of the given equations. y = 1/2x2, y = -x2 + 6

Answer 1

that why i also aske people to double check that they typed the right question

Answer 2

yes, sorry

Answer 3

now the limits of intergration are from -2 tro 2

Answer 4

so, again all you have to do is take the intergral from 0 to 2 of 6-5x^2/2 and double it

Answer 5

now, could you please help me?

Answer 6

well you need to integrate that

Answer 7

is this
\[y=\frac{1}{2}x^2\] adn
\[y=-x^2+6\]

Answer 8

i guess smooth man is typing up a long response, so i will wait for him

Answer 9

-x2+6 = 1/2x2
6 = 3/2x2
4 = x2
x = +-2

integral from -2 to 2 of 6-x2-1/2x2 =
int(-2 to 2) 6-3/2x2

Antiderivative is 6x - 3/6x3

evaluate from -2 to 2. Or from 0 to 2 and then double the result.

6(2) - 1/2(2)^3 - (6(-2) - 1/2(-2)^3) =
12 - 4 - (-12 + 4) = 24 - 8 = 16

Answer 10

Yes, Satellite. Those are the functions.

Answer 11

\[2\int _0^2-x^2+6-\frac{1}{2}x^2dx\] yes?

Answer 12

Yeah. That would give it to you, Satellite.

Answer 13

or somewhat more elegantly
\[2\int_0^2-\frac{3}{2}x^2+6dx\]

Answer 14

\[2\int_0^26dx =2\times 2\times 6=24\] by inspection

Answer 15

Um what.

Answer 16

Oooh. Nevermind. I see.

Answer 17

and
\[2\times \int_0^2 -\frac{3}{2}x^2dx=2\times -\frac{x^3}{2}|_0^2=-2^3=-8\]

Answer 18

Yes. =) So the sum is 16.

Answer 19

so i guess answer is 24-8=16 joui?

Answer 20

ok got it!

Answer 21

thank you. i have cal 2 exam today( it is online) . i have lots of questions . can i ask ?

Answer 22

yes, i don't know why i add 24+ 8 then 32. i solved it more than 10 times but maybe careless.

Answer 23

knock yourself out, but best to post them fresh. and if you are taking an on line exam and want to cheat, here is fine but doing them yourself here
http://www.wolframalpha.com/
pretty much guarantees you get the right result, assuming you are careful with typing.

Answer 24

no. i had 60 questions as home work. i have 20 of them to ask. when i finish with it. i will go and take it myself
i want to learn. i do not want to cheat.

Answer 25

com on just admit you wanna cheat lol

Answer 26

I admire that so hard. If you legit want to learn and understand, you can expect a lot of help from me.

Answer 27

can i ask more questions here?

Answer 28

Yeah absolutely. The thread tends to lag as it gets longer though, so it's probably worth posting a new question.

Answer 29

i have posted two questions but i did not get any feed back. what should i do., i am completely new and i don't know what to do .