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Let $a\in\mathbb{R}$ and $f:[a-1,a+1]\longrightarrow\mathbb{R}$ be a differentiable function, for which $f(a-t)=f(a+t)$ for all t $\in[-1,1]$. Prove, that $$\int_{a-1}^{a+1}xf(x)\ dx=2a\int_a^{a+1}f(x) \ dx\ .$$ I know I need a clever substituion, but all I could think about was $u=a-x$ which didn't get me anywhere. Especially, I'm not sure how to obtain interval that begins in $a$ instead of $a-1$. I could use a hint.

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2 Answers 2

up vote 3 down vote accepted

Applying $\displaystyle \int_a^b f(x) dx = \int_a^b f(a+b-x) dx $ i.e $x \to a+b -x$ on integrand gives $$\int_{a-1}^{a+1}xf(x)\ dx=\int_{a-1}^{a+1} 2a f(2a-x)dx - \int_{a-1}^{a+1} x f(2a-x)dx$$

If you make substition here $f(a-t)=f(a+t)$, $t\to -a+x$ you get $f(2a-x) = f(x)$ where $x \in [a-1, a+1]$.

Or, $$I = 2a \int_{a-1}^{a+1}f(x) dx - I \implies I = a \int_{a-1}^{a+1}f(x) dx$$

Also $$\int_{a-1}^{a+1}f(x) dx = \int_{a-1}^a f(x) dx + \int_a^{a+1}f(x) dx$$ Using the same trick, and making subs $x+1 \to x $ we get $$\int_{a-1}^a f(2a-1-x) dx = \int_{a-1}^a f(1+x)dx = \int_a^{a+1}f(x) dx$$ which is your required answer. $$\int_{a-1}^{a+1}xf(x)\ dx=2a\int_a^{a+1}f(x) \ dx\ $$

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Hint: $\displaystyle\int_{a-1}^{a+1}xf(x)dx=\int_{-1}^{1}(x+a)f(x+a)dx$

Edit: Since someone else has posted a full solution, I thought I might as well flesh the hint out to a full answer too.

$\displaystyle\int_{a-1}^{a+1}xf(x)dx=\int_{-1}^{1}(x+a)f(x+a)dx=\int_0^1(x+a)f(x+a)dx+\int_{-1}^0(x+a)f(x+a)dx\\\displaystyle=\int_0^1(x+a)f(x+a)dx+\int_0^1(-x+a)f(x+a)dx=2a\int_0^1f(x+a)dx=2a\int_{a}^{a+1}f(x)dx$

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