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How can I prove $\int_{-a}^0f(x)dx=\int_0^{a}f(-x)dx$? I can't say anything about the relationship between $f(x)$ and $f(-x)$ without knowing whether it's an even or odd function. The problem says to use a change of variable but I don't understand how I can't change any variables without a function.

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Let's take $y = -x$. Then, $f(-x) = f(y)$, and $dx = -dy$.

Then, $\int_0^a f(-x)\ dx = \int_0^a -f(y)\ dy$. This is by a simple change of variables.

Now, you need to know that in general, $\int_a^b g(t)\ dt = - \int_b^a g(t)\ dt$; in other words, the negative of an integral is the same thing as the integral with the limits swapped. Note that this comes straight from the Fundamental Theorem of Calculus:

$$\int_a^b g(t)\ dt = G(b)-G(a)$$

as $G(a)-G(b) = -(G(b)-G(a))$.

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You could look at it geometrically as well as with a $u$-substitution. The graph of $f(-x)$ is the graph of $f(x)$ rotated $180$ degrees across the $y$-axis. So the area between the $x$-axis and $f(-x)$ from $0$ to $a$ becomes the area between the $x$-axis and $f(x)$ from $-a$ to $0$ upon rotation. This may not be very rigorous, but it helps with the intuition.

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Let $u = -x$ so that $du = -dx$ and perform $u$-substitution on the right integral. Notice that your bounds will change, instead of being from $0$ to $a$, they will be from $0$ to $-a$.

With the $u$-substitution, our integral becomes

$$\int_0^{-a} -f(u)\ du.$$ Of course we can change the order of our bounds by multiplying by a negative,

$$\int_{-a}^0 f(u)\ du.$$

But this is the original integral on the left, $\int_{-a}^0 f(x)\ dx$, since it doesn't matter what dummy variable we use for integration.

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  • $\begingroup$ So if I have $\int_{-a}^0f(x)dx=\int_{-a}^0f(u)du$ what do I do about f(x) and f(u) being different? $\endgroup$
    – user215498
    Feb 14, 2015 at 22:14
  • $\begingroup$ Note that when you change variable from $x$ to $u$, you are also changing indexes. In particular $u = -x$ implies $-u = x$ so $ x = 0 \Rightarrow u = -0$ and $ x = -a \Rightarrow u = -(-a) = a$ $\endgroup$
    – Blex
    Feb 14, 2015 at 22:20

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