Question

graph y=4x-x^3. Using integration find the area bound by the curve and the x axis. THANK YOU

Answer 1

So the first thing you do when trying to find the area bounded by a curve is to find the intersections, in this case with the x axis.

Answer 2

i think they are 0 -2 and 0 2?

Answer 3

oops
you're right

Answer 4

\[\frac{4x^{1+1}}{2}-\frac{x^{3+1}}{3+1}=2x^2-\frac{x^4}{4}\]
you will use this

Answer 5

yes you're right the intersections are 0,2 and 0,-2

Answer 6

ok

Answer 7

so now you have the bounds so you integrate from -2 to 2

Answer 8

wrong

Answer 9

\[4x-x^3=x(4-x^2)=x(2-x)(2+x)=0 =>x=0,x=2,x=-2\]

Answer 10

so i just plug in 2 into the formula then plug in -2 and subtract those two answers?

Answer 11

the integral will be zero , because its an odd function

Answer 12

but the area wont be zero

Answer 13

the answer that my book says is 0..

Answer 14

yeh well they are wrong , if they asked for area

Answer 15

\[\int\limits_{-2}^{2}4x-x^3dx\]
is equal to
\[2x^2-1/4x^4 from -2 \to 2\]
is equal to 4-4 which would equal 0

Answer 16

i still get turned around on what they really want in thos

Answer 17

you need to find the integral from 0 to 2

Answer 18

however elecengineer is correct

Answer 19

then double it

Answer 20

\[\int\limits_{-2}^{0}f(x)dx+ \int\limits_{0}^{2}f(x)dx\]

Answer 21

there is an area of 4 before x=0 and an area of 4 after x=0 which =8

Answer 22

^ still one of them needs absolute value, the first one

Answer 23

so the answer is 0 or not?

Answer 24

if they asked for "integral" then yes

Answer 25

if they asked for "area" then no

Answer 26

integral or net area*

Answer 27

elecengineer is right
if they asked for the integral (which they didn't) then the answer is 0
if they asked for the area (which they did) then the answer is 8

Answer 28

Thanks everyone

Answer 29

np