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anonymous
 one year ago
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
 one year ago
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 ...

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UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3ok... so we are given a function \[\LARGE f(x) = x^3\] graphed between the intervals of 2 to 2

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3and then we are asked how the plot of that graph gave away the value when we are integrating from 2 to 2

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3you wouldn't mind if we integrate first right? do you know antiderivatives?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Hint : if \(f(x)\) is an odd function, \[\int\limits_{a}^a f(x)\,dx = 0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I dont mind, but negative on the anti derivatives (I think)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3when we are taking the antiderivative, we add one to the exponent and divide by the new exponent.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so since your exponent is 3. What's 3 +1 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so do they mean, I am to look at the plot, and then work out the value of the integration equation?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1436955882543:dw now evaluate f(b)f(a) or f(2)f(2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when 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
 one year ago
Best ResponseYou've already chosen the best response.1Notice 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
 one year ago
Best ResponseYou've already chosen the best response.0oh, 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
 one year ago
Best ResponseYou've already chosen the best response.3yes 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
 one year ago
Best ResponseYou've already chosen the best response.3but keep in mind we are restricted from 2 to 2 as well

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you both.. gotcha, I just wasn't sure why they would ask me about.. 'giving away' .. I was like.. "giving away what?"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The author of this program uses so many strange metaphors to me

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1436956101513:dw well the result does come to 0 if we do evaluate after taking the antiderivative.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so there is cancellation...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Indeed, the graph does give away the fact the the given function is "odd" dw:1436956075476:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so 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 antiderivative and evaluating... we will have 0 (though I like to work backwards....easier)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3maybe the cancellations gave out the answer to that integral ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1You 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
 one year ago
Best ResponseYou've already chosen the best response.0gotcha, 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..

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3another odd function... thanks to symmetry on the origin

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1same thing, just check if it is an odd one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm, they might want me to do it analytically too..

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I don't think we need integration by parts on this one... just usubstitution.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436956504186:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well there is that f[x] == f[x] symmetry going on.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, you must be telepathic

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you for your instant help guys.. I dont think it was 30 seconds and you had me covered.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0please someone approve usukiDoll for qualified helper status.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I was going to do my nails when I saw the QH Questions change color, so I was like someone just asked a question.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3Good thing I didn't have wet polish otherwise I would have to do it over XD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do 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
 one year ago
Best ResponseYou've already chosen the best response.3if 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
 one year ago
Best ResponseYou've already chosen the best response.3there's also a test too.. that determines if the function is odd or not

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3now if we have an even function, that means that there is symmetry on the yaxis the parabola \[\large f(x) =x^2\] is the most common example for an even function

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3for trigonometry , sinx cosx, tanx sinx and tanx are odd functions cosx is the only even function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So then it is only functions then that exhibit the property of f[x] + f[x] == 0 that are considered odd?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1For 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
 one year ago
Best ResponseYou've already chosen the best response.3erm.. you have to plug in x in your x and yes the definitions of odd and even functions help a lot.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3So, 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
 one year ago
Best ResponseYou've already chosen the best response.1for example below graph is odd too; there is no algebraic expression to express it though dw:1436958097638:dw