I can't believe sometimes that English is my first language. What does this question mean?
"Here is a plot of f[x] = x^3 on [-2,2]
How does the plot give away the value of Integrate[x^3, {x, -2, 2}] ?"
pictures coming ...

- anonymous

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- chestercat

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- anonymous

image attached

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- UsukiDoll

ok... so
we are given a function
\[\LARGE f(x) = x^3\] graphed between the intervals of -2 to 2

- anonymous

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- UsukiDoll

and then we are asked how the plot of that graph gave away the value when we are integrating from -2 to 2

- UsukiDoll

you wouldn't mind if we integrate first right? do you know anti-derivatives?

- ganeshie8

Hint : if \(f(x)\) is an odd function,
\[\int\limits_{-a}^a f(x)\,dx = 0\]

- anonymous

I dont mind, but negative on the anti derivatives (I think)

- UsukiDoll

when we are taking the antiderivative, we add one to the exponent and divide by the new exponent.

- UsukiDoll

so since your exponent is 3. What's 3 +1 ?

- anonymous

so do they mean, I am to look at the plot, and then work out the value of the integration equation?

- anonymous

hmmm.. 5-1?

- anonymous

j/k 4

- UsukiDoll

|dw:1436955882543:dw|
now evaluate
f(b)-f(a)
or f(2)-f(-2)

- anonymous

when the question says.. the value of ... do they mean that the equation itself is a value, or that the equation comes out to some value?

- ganeshie8

Notice that the graph is symmetric about origin, so f(x) is an odd function. You don't need to do any further work, the definite integral between a symmetric interval would be simply 0 because the positive area and negative area kill each other out.

- anonymous

oh, so by looking at the image.. I should see that it is symmetrical about zero, and therefore the sum of it's parts will be a negative area + an equal positive area = zero.

- UsukiDoll

yes it is an odd function, so we do have symmetry around the origin . When we graph x^3 only we should have just |dw:1436956009479:dw|

- UsukiDoll

but keep in mind we are restricted from -2 to 2 as well

- ganeshie8

Exactly!

- anonymous

thank you both.. gotcha, I just wasn't sure why they would ask me about.. 'giving away' .. I was like.. "giving away what?"

- anonymous

The author of this program uses so many strange metaphors to me

- UsukiDoll

|dw:1436956101513:dw|
well the result does come to 0 if we do evaluate after taking the anti-derivative.

- UsukiDoll

so there is cancellation...

- ganeshie8

Indeed, the graph does give away the fact the the given function is "odd"
|dw:1436956075476:dw|

- UsukiDoll

so by having an odd function, there's symmetry around the origin, the negative and positive cancel out and that leads to a 0
which is the same answer if we were able to evaluate that integral.
By taking the anti-derivative and evaluating... we will have 0
(though I like to work backwards....easier)

- UsukiDoll

maybe the cancellations gave out the answer to that integral ?

- ganeshie8

You could also work it analytically :
\(f(-x) = (-x)^3 = -x^3 = -f(x)\)
\(f(-x) = -f(x)\) means \(f(x)\) is an odd function.

- anonymous

gotcha, I haven't got to anti derivatives yet.. I think they're coming up.. and they just want me to look at the graph.. there's another identical one next..

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- UsukiDoll

another odd function... thanks to symmetry on the origin

- ganeshie8

same thing, just check if it is an odd one

- anonymous

hmmm, they might want me to do it analytically too..

- UsukiDoll

I don't think we need integration by parts on this one... just u-substitution.

- ganeshie8

|dw:1436956504186:dw|

- anonymous

Well there is that f[-x] == -f[x] symmetry going on.

- anonymous

yeah, you must be telepathic

- UsukiDoll

xD!

- anonymous

thank you for your instant help guys.. I dont think it was 30 seconds and you had me covered.

- anonymous

please someone approve usukiDoll for qualified helper status.

- UsukiDoll

I was going to do my nails when I saw the QH Questions change color, so I was like someone just asked a question.

- UsukiDoll

Good thing I didn't have wet polish otherwise I would have to do it over XD

- anonymous

Do all odd functions center on origin?
or is this only because x^3 is fairly basic.
And by odd did you mean the expression had degree 3, or odd means any function of an odd degree?

- UsukiDoll

if the function is odd then there is symmetry at the origin
and yes we need an odd exponent number like
\[\large f(x) = x^3,f(x)=x^5.f(x)=x^7\]

- UsukiDoll

there's also a test too.. that determines if the function is odd or not

- UsukiDoll

now if we have an even function, that means that there is symmetry on the y-axis
the parabola \[\large f(x) =x^2\] is the most common example for an even function

- UsukiDoll

for trigonometry , sinx cosx, tanx
sinx and tanx are odd functions
cosx is the only even function

- anonymous

So then it is only functions then that exhibit the property of
-f[x] + f[x] == 0 that are considered odd?

- ganeshie8

For simple polynomials we do need odd degree, but that is not so important.
It would be more simple to just stick to the definition :
\(f(x)\) is an odd function if \(f(-x) = -f(x)\)

- UsukiDoll

erm.. you have to plug in -x in your x
and yes the definitions of odd and even functions help a lot.

- UsukiDoll

So, take the function and plug in -x for x and simplify
should our result become
\[\large f(-x)= f(x) \]
the function is even and are signs are the same
If our signs are the opposite (for example minus signs become plus signs or plus signs become minus signs) then we have an odd function
\[\large f(-x)=-f(x) \]
There is a chance that our function is neither odd nor even

- ganeshie8

for example below graph is odd too; there is no algebraic expression to express it though
|dw:1436958097638:dw|