Монгол Бодлогын Сан

Заавар: $$\int_a^b f(\varphi(t))\cdot\varphi^\prime(t)\,\mathrm{d}t=\int_{\varphi(a)}^{\varphi(b)} f(x)\,\mathrm{d}x$$

Бодолт: $t=\sqrt{1+x}$ орлуулга ашиглавал $$\begin{align*} \int_0^1\sqrt{1+x}\,\mathrm{d}x&=\left[\begin{array}{c}t=\sqrt{1+x}\\ x=t^2-1\\ \mathrm{d}x=2t\,\mathrm{d}t \end{array} \right]\\ &=\int_{\sqrt{1+0}}^{\sqrt{1+1}}t\cdot 2t\,\mathrm{d}t\\ &=\int_{1}^{\sqrt2}2t^2\,\mathrm{d}t\\ &=\dfrac{2t^3}{3}\bigg|_{1}^{\sqrt2}\\ &=\dfrac{2}{3}((\sqrt{2})^3-1^3)\\ &=\dfrac23(\sqrt{8}-1) \end{align*}$$