Монгол Бодлогын Сан

Заавар:

Бодолт: $5x-3=0$ гэвэл $x=3/5$ байна. $$\displaystyle\int_{-2}^{1}|5x-3|\,\mathrm{d}x=\int_{-2}^{\frac35}|5x-3|\,\mathrm{d}x+\int_{\frac35}^1|5x-3|\,\mathrm{d}x=$$ $$\displaystyle=-\int_{-2}^{\frac35} 5x-3 \,\mathrm{d}x+\int_{\frac35}^1 5x-3 \,\mathrm{d}x=-\big(\tfrac52x^2-3x\big)\Big|_{-2}^{\tfrac35}+\big(\tfrac52x^2-3x\big)\Big|_{\frac35}^{1}=$$ $$=-\big(\tfrac52\cdot\big(\tfrac35\big)^2-3\cdot\tfrac35\big)+\big(\tfrac52\cdot(-2)^2-3\cdot(-2)\big)+\big(\tfrac52\cdot1^2-3\cdot1\big)-\big(\tfrac52\cdot\big(\tfrac35\big)^2-3\cdot\tfrac35\big)=$$ $$=-\big(\tfrac{9}{10}-\tfrac{9}{5}\big)+(10+6)+\big(\tfrac52-3\big)-\big(\tfrac{9}{10}-\tfrac{9}{5}\big)=\tfrac{9}{10}+16-\tfrac12+\tfrac{9}{10}=17\tfrac{3}{10}.$$

Тодорхой интегралын геометр утгыг ашиглан бодож болно. $A(-2;13)$, $B(-2;0)$, $C(3/5;0)$, $D(1;0)$, $E(1;2)$ гэвэл $$\int_{-2}^{1}|5x-3|\,\mathrm{d}x=S_{\triangle ABC}+S_{\triangle CDE}=\tfrac{13\cdot(3/5+2)}{2}+\tfrac{(1-3/5)\cdot 2}{2}=17\tfrac{3}{10}$$ байна.