Монгол Бодлогын Сан

Заавар: $(1) \displaystyle\int_{\alpha}^{\beta}(x-\alpha)^2\,dx=\left.\dfrac{(x-\alpha)^3}{3}\right|_{\alpha}^{\beta}= \dfrac{(\beta-\alpha)^3}{3}$

$(2)$ $$\begin{align*} (x-\alpha)(x-\beta)&=(x-\alpha)((x-\alpha)+(\alpha-\beta))\\ &=(x-\alpha)^2+(\alpha-\beta)(x-\alpha) \end{align*}$$ тул $$\begin{align*} \int_{\alpha}^{\beta}(x-\alpha)(x-\beta)\,dx&=\int_{\alpha}^{\beta}(x-\alpha)^2\,dx+(\alpha-\beta)\int_{\alpha}^{\beta}(x-\alpha)\,dx\\ &=\left.\left\{\dfrac{(x-\alpha)^3}{3}+(\alpha-\beta)\cdot\dfrac{(x-\alpha)^2}{2}\right\}\right|_{\alpha}^{\beta}\\ &=\dfrac{(\beta-\alpha)^3}{3}-\dfrac{(\beta-\alpha)^3}{2}=-\dfrac{(\beta-\alpha)^3}{6} \end{align*}$$

Бодолт:

1. $\int\limits_{2}^{5}(x-2)^2\,dx=\dfrac13(5-2)^3=9$.
   
   
2. $x^2-2x-1=0$ тэгшитгэлийн шийдүүд $\alpha=1-\sqrt{2}$, $\beta=1+\sqrt{2}$ тул $$2\int_{\alpha}^{\beta}(x^2-2x-1)\,dx=2\cdot \left(-\dfrac16\right)(\beta-\alpha)^3=-\dfrac{16\sqrt{2}}{3}$$
   
   
3. $\alpha=\dfrac{\sqrt{5}-1}{2}$ гэвэл $\alpha^2+\alpha-1=0$ тул $$\begin{align*} \int_{0}^{\alpha}(x^2+x-1)\,dx&=\dfrac{\alpha^3}{3}+\dfrac{\alpha^2}{2}-\alpha=\dfrac16(2\alpha^3+3\alpha^2-6\alpha)\\ &=\dfrac16(2\alpha(\alpha^2+\alpha-1)+(\alpha^2+\alpha-1)-5\alpha+1)\\ &=\dfrac{1-5\alpha}{6}=\dfrac{7-5\sqrt{5}}{12} \end{align*}$$