回復 12# 阿光 的帖子
計算2.
(1)已知\(ax^2+bx+c=0\)的兩實根為\(\alpha\)與\(\beta\),試證:\( \displaystyle \int_{\alpha}^{\beta}(ax^2+bx+c)dx=\frac{1}{6}a(\alpha-\beta)^3 \)。
[解答]
沒事找事用怪招
令 \( f(x) = x(x-1) \),易知 \( \int_0^1 f(x) dx = -\frac16 \)
設 \( \alpha -\beta \neq 0 \),則 \( ax^{2}+bx+c=a(\alpha-\beta)^{2}f(\frac{x-\alpha}{\beta-\alpha}) \)
故 \( \int_\alpha^\beta ax^2+bx+c dx = \int_{\alpha}^{\beta}a(\alpha-\beta)^{2}f(\frac{x-\alpha}{\beta-\alpha})dx\)
變數代換令 \( \displaystyle t = \frac{x-\alpha}{\beta-\alpha} \),則上式又可整理為
\( \displaystyle \int_{\alpha}^{\beta}a(\alpha-\beta)^{2}f(\frac{x-\alpha}{\beta-\alpha})dx=a(\beta-\alpha)^{3}\int_{0}^{1}f(t)dt=a(\alpha-\beta)^{3} \)
實際上,就是做了伸縮變換,面積常數倍,把常數寫下來,左右 \( \beta -\alpha \) 倍,上下 \( a(\alpha-\beta)^2 \) 倍