dude, in my opinion, unless the function is symmetric about the origin, you cannot change the limits from -1...1 to 2(0..1)

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Originally posted by twilightki : It feels like something you'd listen to at 4 in the morning, or listen to in your car while you're going in a tunnel.

the conclusion that ive come to is that Equation 1 and equation 2 must be different hence why i get a different solution. But logically if you half the range and multiply it by 2 you should get the same area by going from -1 to 1

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Originally posted by Omega_M dude, in my opinion, unless the function is symmetric about the origin, you cannot change the limits from -1...1 to 2(0..1)

it is symmetric about the origin.


it looks like this /\ where the peak is at the origin. aneven function with the even side at 2+x (left) and the right side at 2-x

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Originally posted by Nrg2Nfinit it is symmetric about the origin. it looks like this /\ where the peak is at the origin. aneven function with the even side at 2+x (left) and the right side at 2-x

2+x on left and 2-x on right means they are 2 different functions.

y = 2-x is not symmetric about origin.

for x=-1, y = 3
for x = 1, Y = 1

For symmetry, the function Y(-1) = Y(1).

You need to split the integral from -1 to 0 and 0 to 1 and then solve. Between - 1 to 0, use 2+x, between 0 and 1, use 2-x. The answer is 3.

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Originally posted by twilightki : It feels like something you'd listen to at 4 in the morning, or listen to in your car while you're going in a tunnel.

No ?

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Originally posted by twilightki : It feels like something you'd listen to at 4 in the morning, or listen to in your car while you're going in a tunnel.

The corret answer = NO VAGINA.

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Mind your own business!

2+x and 2-X is symmetric about the y-axis.

but do you not see they are 2 different functions ? Hence you need to split the integral as I did..and I'm getting 3, which is the answer in your book.

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Originally posted by twilightki : It feels like something you'd listen to at 4 in the morning, or listen to in your car while you're going in a tunnel.

quote:

Originally posted by Omega_M but do you not see they are 2 different functions ? Hence you need to split the integral as I did..and I'm getting 3, which is the answer in your book.

yeah i can see that

i guess i dont understand hte concept of even and odd functions. If a function is even can we automatically half the integral range and double it?

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Originally posted by Nrg2Nfinit yeah i can see that i guess i dont understand hte concept of even and odd functions. If a function is even can we automatically half the integral range and double it?

For the function y(x) to be even, the condition y(x)= y(-x)must be satisfied. The function needs to be continuous. Like cos(x). You can check that cos(x) = cos(-x). Hence it's even symmetric. If you look at the equation you provided, it really is 2 different functions, not one. Hence it's neither even nor odd. But if it is even then yes, you can half the range and double the integral.

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Originally posted by twilightki : It feels like something you'd listen to at 4 in the morning, or listen to in your car while you're going in a tunnel.

quote:

Originally posted by Omega_M For the function y(x) to be even, the condition y(x)= y(-x)must be satisfied. The function needs to be continuous. Like cos(x). You can check that cos(x) = cos(-x). Hence it's even symmetric. If you look at the equation you provided, it really is 2 different functions, not one. Hence it's neither even nor odd. But if it is even then yes, you can half the range and double the integral.

here it is. Question 1)

http://math.carleton.ca:16080/~math..._test1a_sol.pdf


it states that the function is even.