by infinite-blank-
Last Updated July 12, 2019 07:20 AM

I have to calculate this integral: $$\int_{-10}^{10}\frac{3^x}{3^{\lfloor x\rfloor}}dx$$

I know that this function ie. $$3^{x-\lfloor x\rfloor}$$ is periodic with period $T=1$ so I rewrote the integral as $$20\int_{0}^{1}\frac{3^x}{3^{\lfloor x\rfloor}}dx$$

But the problem is that I can't figure out how to calculate the final integral.

Any help is appreciated.

**Hint**

$$\int_{-10}^{10} \frac{3^x}{3^{\lfloor x\rfloor }}\,\mathrm d x=\sum_{i=-9}^{10}\frac{1}{3^{k-1}}\int_{k-1}^k 3^x\,\mathrm d x.$$

Updated September 30, 2017 10:20 AM

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