Let $f$ be continous in $[-2,0]$ and $\displaystyle\int\limits_{-2}^0 f(x) \,\text{d}x=3 $. I'm supose to prove that $\displaystyle\int\limits_0^2 f(x-2) \,\text{d}x =3 $ but without $u$ substitutions. How to prove it ?
2026-04-10 12:10:41.1775823041
Given that $\int\limits_{-2}^0 f(x) \,\text{d}x=3 $, show that $\int\limits_0^2 f(x-2) \,\text{d}x =3 $.
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Use the Second Fundamental Theorem of Calculus namely:
$\int_a^bf(x)dx=F(b)-F(a)$