Is $\int\limits_{-a}^{0} f(x)\,\mathrm{d}x = \int\limits_0^a f(-x)\,\mathrm{d}x$ true?

Question

Is the following theorem true? If yes, how to prove it?

$$\int\limits_{-a}^{0} f(x)\,\mathrm{d}x = \int\limits_0^a f(-x)\,\mathrm{d}x$$

Update: $$ \int\limits_{\color{red}-a}^{0} f(x) \,\mathrm{d}x = \color{red}-\int\limits_{\color{red}+a}^{0} f(\color{red}-x) \,\mathrm{d}x = \int\limits_{0}^{\color{red}+a} f(\color{red}-x)\,\mathrm{d}x $$

does that make sense?

score 4 · Answer 1 · answered Feb 20 '13 at 00:04

4

It is true, to prove it try doing a $u$-substitution with $u = -x$.

answered Feb 20 '13 at 00:04

Jim

30,682

score 2 · Answer 2 · answered Feb 20 '13 at 00:20

2

Define $F(x): \frac{d}{dx} F(x) = f(x)$.

Then the first integral evaluates to $F(0) - F(-a)$.

Now for the second integral use u = -x (as Jim said). Then du = -dx.

We also have to convert the limits of integration. $x=0\Rightarrow u = 0, x=a\Rightarrow u = -a$.

And we have:

-$\int\limits_0^{-a}f(u)du = -(F(-a)-F(0)) = F(0)-F(-a) = \text{the first integral. There you have it.}$

answered Feb 20 '13 at 00:20

daniel.wright

303

Thanks for your answer, but I would like to have an $a$ in the integral, instead of the $-a$. – sergej Feb 20 '13 at 09:16
Why does $f$ have anti-derivative? – Lior B-S Feb 20 '13 at 10:10

Is $\int\limits_{-a}^{0} f(x)\,\mathrm{d}x = \int\limits_0^a f(-x)\,\mathrm{d}x$ true?

2 Answers2