# Help understanding an integral rule

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• July 24th 2009, 03:55 AM
kallekall
Help understanding an integral rule
Hi again :)

I wonder why

$\int_{-\pi}^{0} |\varphi|\sqrt{\varphi^2 +4}d\varphi = \int_{0}^{\pi} \varphi\sqrt{\varphi^2 +4}d\varphi$

Since one are suppose to always submit the whole problem:
$r=\varphi^2, -\pi\leq\varphi\leq\pi$ Calculate curvelength.

I used: $ds=\int_{-\pi}^{\pi} \sqrt{\varphi^4 +(2\varphi)^2}d\varphi$
• July 24th 2009, 04:16 AM
Moo
Hello,

Substitute $t=-\varphi$

And you'll get the new integral :)
• July 24th 2009, 05:27 AM
kallekall
I didn't really manage to use your method but I called a friend and he said that curvelenght/area for $\int_{-a}^0$ always is equal to $\int_0^{a}$ so I guess that I will just remeber that rule. But is that really true? What for $\int_{-1,9}^0\frac{dx}{x-2}$
• July 24th 2009, 06:07 AM
Plato
Quote:

Originally Posted by kallekall
curvelenght/area for $\int_{-a}^0$ always is equal to $\int_0^{a}$ so I guess that I will just remeber that rule. But is that really true? ]

Well of course that is not true.

But notice that $f(x) = \left| \varphi \right|\sqrt {\varphi ^2 + 4}$ is an even function.

Then for any even function we have $\int\limits_{ - a}^a f = 2\int\limits_0^a f$
• July 24th 2009, 07:41 AM
HallsofIvy
Moral, do not use rules you do not understand!(Happy)

If f(x) is an even function, that is f(-x)= f(x), then $\int_{-a}^0 f(x) dx= \int_a^0 f(-u)(-du)= \int_0^a f(u)du$, using the substitution u= -x.

So, again for an even function, $\int_{-a}^a f(x)dx= \int_{-a}^0 f(x)dx+ \int_0^a f(x)dx= \int_0^a f(u)du+ \int_0^a f(x)dx= 2\int_0^a f(x)dx$

If f(x) is an odd function, that is f(-x)= -f(x), then $\int_{-a}^0 f(x) dx= \int_a^0 -f(-u)(-du)= -\int_0^a f(u)du$, using the substitution u= -x.

Then, for an odd function, $\int_{-a}^a f(x)dx= -\int_{0}^a f(x)dx+ \int_0^a f(x) dx= 0$.

If f(x) is neither even nor odd, neither of those is true.
• July 25th 2009, 03:50 AM
kallekall
Thanks guys! You have helped me a lot!