Diketahui fungsi \( f(x) = f(x+2) \) untuk setiap \(x\). Jika \( \int_0^2 f(x) \ dx = B \), maka \( \int_3^7 f(x+8) \ dx = \cdots \)
- B
- 2B
- 3B
- 4B
- 5B
(SBMPTN 2016)
Pembahasan:
Ingat bahwa jika \( f(x) = f(x+c) \) maka \( f(x) \) adalah fungsi periodik dengan periode \(c\) sehingga berlaku:
\begin{aligned} \int_a^b f(x) \ dx = \int_{a+c}^{b+c} f(x) \ dx = \int_{a+2c}^{b+2c} f(x) \ dx = \cdots \\[8pt] \int_a^b f(x) \ dx = \int_a^b f(x+c) \ dx = \int_a^b f(x+2c) \ dx = \cdots \end{aligned}
Berdasarkan hasil di atas, diperoleh:
\begin{aligned} \int_3^7 f(x+8) \ dx &= \int_3^7 f(x) \ dx \\[8pt] &= \int_3^4 f(x) \ dx + \int_4^6 f(x) \ dx + \int_6^7 f(x) \ dx \\[8pt] &= \int_{3-2}^{4-2} f(x) \ dx + \int_{4- 2 \cdot 2}^{6-2\cdot 2} f(x) \ dx + \int_{6-3 \cdot 2}^{7-3 \cdot 2} f(x) \ dx \\[8pt] &= \int_1^2 f(x) \ dx + \int_0^2 f(x) \ dx + \int_0^1 f(x) \ dx \\[8pt] &= \int_0^1 f(x) \ dx + \int_1^2 f(x) \ dx + \int_0^2 f(x) \ dx \\[8pt] &= \int_0^2 f(x) \ dx + \int_0^2 f(x) \ dx \\[8pt] &= B+B \\[8pt] &= 2B \end{aligned}
Jawaban B.