Монгол Бодлогын Сан

Заавар: $x=a$ цэгээр хоёр хэсэг салгаж бод.

Бодолт: $$\begin{align*} \int_{0}^{2a}x|x^2-a^2|\,\mathrm{d}x&=\int_{0}^{a}x|x^2-a^2|\,\mathrm{d}x+\int_{a}^{2a}x|x^2-a^2|\,\mathrm{d}x\\ &=\int_0^a x(a^2-x^2)\,\mathrm{d}x+\int_0^a x(x^2-a^2)\,\mathrm{d}x\\ &=\int_0^a a^2x-x^3\,\mathrm{d}x+\int_0^a x^3-a^2x\,\mathrm{d}x\\ &=\left(\dfrac{a^2x^2}{2}-\dfrac{x^4}{4}\right)\Bigg|_0^a+\left(\dfrac{x^4}{4}-\dfrac{a^2x^2}{2}\right)\Bigg|_a^{2a}\\ &=\left(\dfrac{a^4}{2}-\dfrac{a^4}{4}-0\right)+\left(\dfrac{16a^4}{4}-\dfrac{4a^4}{2}-\dfrac{a^4}{4}+\dfrac{a^4}{2}\right)\\ &=\dfrac{a^4}{4}+\dfrac{9a^4}{4}=\dfrac{5}{2}a^4 \end{align*}$$