Монгол Бодлогын Сан
Эх хэлээрээ суралцаж, эх хэлээрээ мэдлэгээ түгээе.
Тодорхой интеграл бодох (3)
$(1)$, $(2)$-ийг батлаад түүнийгээ ашиглан $1\sim3$-ийг бод. \begin{align*} &\int_{\alpha}^{\beta}(x-\alpha)^2\,dx=\dfrac13(\beta-\alpha)^3 & (1)\\ &\int_{\alpha}^{\beta}(x-\alpha)(x-\beta)\,dx=-\dfrac{1}{6}(\beta-\alpha)^3 & (2) \end{align*}
- $\displaystyle\int_{2}^{5}(x-2)^2\,dx$
- $\displaystyle\int_{1-\sqrt{2}}^{1+\sqrt{2}}(2x^2-4x-2)\,dx$
- $\displaystyle\int_{0}^{\frac{\sqrt{5}-1}{2}}(x^2+x-1)\,dx$
Бодлогын төрөл: Уламжлалт
Бодлогыг оруулсан: Балхүүгийн Батбаясгалан
Бодолт
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*}
$(2)$ \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:
- $\int\limits_{2}^{5}(x-2)^2\,dx=\dfrac13(5-2)^3=9$.
- $x^2-2x-1=0$ тэгшитгэлийн шийдүүд
$\alpha=1-\sqrt{2}$, $\beta=1+\sqrt{2}$ тул
$$2\int_{\alpha}^{\beta}(x^2-2x-1)\,dx=2\cdot
\left(-\dfrac16\right)(\beta-\alpha)^3=-\dfrac{16\sqrt{2}}{3}$$
- $\alpha=\dfrac{\sqrt{5}-1}{2}$ 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*}