periodic Integration problem...

[Today123]

Suppose \(\displaystyle f\,:\, \mathbb{R}\, \rightarrow \, \mathbb{R}\) is an integrable function satisfying the following:

. . . . .\(\displaystyle \int_{-2}^3\, f(x)\, dx\, =\, 11\) and \(\displaystyle \int_{-2}^0\, f(x)\, dx\, =\, 9\)

Given that \(\displaystyle f\) is a periodic function, and \(\displaystyle f(x)\, =\, f(x\, +\, 2)\) is always true, find:

. . . . .\(\displaystyle \int_2^3\, f(x)\, dx\)

My work: So \(\displaystyle \int_{-2}^0\, f(x)\, dx\, =\, 9\)

Since it is a periodic function then

[image] = -9

That means

\(\displaystyle \int_2^3\, f(x)\, dx\, =\, 11\)

but it is the incorrect answer, any help would be appreciated, I'm not picking something up

 

[HallsofIvy]

Since f is a periodic function with period 2, f(x+ 2)= f(x), and \(\displaystyle \int_{-2}^0 f(t)dt= 9\), then \(\displaystyle \int_0^2 f(t)dt= 9\) also and \(\displaystyle \int_{-2}^2 f(t)dt= 9+ 9= 18\). Since \(\displaystyle \int_{-2}^3 f(t)dt= 11\), \(\displaystyle \int_2^3 f(t)dt= \int_{-2}^3 f(t)dt- \int_{-2}^2 f(t)dt\). That looks like what you said you did but 11- 18 is NOT equal to 11!