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How to prove $\displaystyle \int_a^b g(x) \, dx = \int_a^b g(a+b-x) \,dx$ ?

From where to start proving?

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    Did you try substitution? – 2017-02-08
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    If i start from the right side of equation i could use $ t = a +b -x $, but i will end up with $t$ instead of $x$. I don't see what can i use if i start from the left side. Recomendation? :) – 2017-02-08
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    $x$ is a dummy variable, you can name it like you want. – 2017-02-08

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Hint. Let $t=a+b-x$ then $$\int_a^b g(x)\, dx = \int_{b}^{a} g(a+b-t)\,d(a+b-t).$$

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    OK! This works if i start from the left side of the equation. Is that what you want to show? – 2017-02-08
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    @Ognjen Mojovic Yes, I started from the left side. – 2017-02-08
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Let $v = a+b-x$, $\frac{dv}{dx} = -1$

\begin{align} \int_a^b g(a+b-x) dx = -\int_{a+b-a}^{a+b-b} g(v) dv = -\int_{b}^{a} g(v) dv=\int_{a}^{b} g(v) dv \end{align}

Response to your comment regarding $t$ and $x$, these are just symbols, that is

$$\int_{a}^{b} g(t) dt = \int_{a}^{b} g(x) dx$$

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    I understand this. Thank you! I'm now interested to prove it from the other side of the equation. – 2017-02-08
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    Is this correct: $ \int_a^b f(x)dx = - \int_b^a f(x)dx = | a+b-t=x, dt=-dx, x(b)=a, x(a)=b | = \int_b^a f(x)(-dx) = \int_a^b f(a+b-t)dt = \int_a^b f(a+b-x)dx $ – 2017-02-08
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For a continuous function $f$, $\int_a^b f(x) dx$ is the surface of the area between the $x$-axis, the vertical lines $x=a$ and $x=b$, and the curve of the function $f$.

That being said, you can observe that the curve $(1)$ of $x \mapsto g(a+b-x)$ in the symmetrical of the curve $(2)$ of $x \mapsto g(x)$ with respect to the vertical line $x= \frac{a+b}{2}$. If you can't convince yourself of that geometrically, you can prove it using the following:

Any point $A$ belonging to the curve $(1)$ can be parametrised by coordinates of the form $A(z)=(z, g(a+b-z))$ with $z$ ranging from $a$ to $b$. Any point $B$ belonging to the curve $(2)$ can be parametrised by coordinates of the form $(z, g(z)))$. The mapping $z \mapsto a+b-z$ being bijective from $[a,b]$ to $[a,b]$, these points $B$ can be equivalently parametrised by coordinates of the form $B(z)=(a+b-z, g(a+b-z))$ with $z$ ranging from $a$ to $b$. With this new parametrisation of $B(z)$ shows us that $A(z)$ and $B(z)$ have the same $y$ coordinate, and the average of their $x$ coordinate is exactly $x= \frac{a+b}{2}$; and this, for all $z$ in $[a,b]$. This suffices to proof the axial symmetry we are seeking.

That being said, and given that the verical lines $x=a$ and $x=b$ are the images of each other by the same axial symmetry, and that the $x-$ axis is the image of itself by this axial symmetry (it is orthogonal to the vertical line $x= \frac{a+b}{2}$), we conclude that "the area between the $x$-axis, the vertical lines $x=a$ and $x=b$, and the curve of the function $x \mapsto g(x)$" and "the area between the $x$-axis, the vertical lines $x=a$ and $x=b$, and the curve of the function $x \mapsto g(a+b-x)$" are symmetric to each other.

The symmetry preserves the surface, hence the equality of the integrals $\int_a^b g(x) dx$ and $\int_a^b g(a+b-x) dx$