回覆 7# swallow7103 的帖子
填充8.
求\(\displaystyle \int_{-1}^1 \frac{x^2}{1+2^x}dx=\) 。
[解答]
令 \(t = -x\)，則 \(dt = -dx\)
\(\displaystyle \int_{-1}^1 \frac{x^2}{1+2^x} dx = \int_{1}^{-1} - \frac{t^2}{1+2^{-t}} dt = \int_{1}^{-1} - \frac{2^t \cdot t^2}{2^t+1} dt = \int_{-1}^{1} \frac{2^t \cdot t^2}{2^t+1} dt = \int_{-1}^{1} \frac{2^x \cdot x^2}{2^x+1} dx\)
再由
\(\displaystyle \int_{-1}^1 \frac{x^2}{1+2^x} dx + \int_{-1}^{1} \frac{2^x \cdot x^2}{2^x+1} dx = \int_{-1}^{1} x^2 dx = \frac{2}{3}\)
得 \(\displaystyle \int_{-1}^1 \frac{x^2}{1+2^x} dx = \frac{1}{3}\) 。
註： 其實就是您開頭的第一行第一個等號。
\(\displaystyle \int_{-1}^1 \frac{x^2}{1+2^x} dx = \int_{-1}^1 \frac{(1+2^x)x^2 - 2^x \cdot x^2}{1+2^x} dx =\int_{-1}^1 x^2 dx - \int_{-1}^1 \frac{2^x \cdot x^2}{1+2^x} dx \)
\(\displaystyle= \frac{2}{3} - \int_{-1}^1 \frac{ x^2}{2^{-x}+1} dx \frac{2}{3} - \int_{-1}^1 \frac{x^2}{2^x+1} dx\)
