📝 题目
解 如果求出 $\sqrt{1 - {x}^{2}}$ 的原函数
$$ \frac{1}{2}\arcsin x + \frac{x}{2}\sqrt{1 - {x}^{2}}, $$
再利用牛顿-莱布尼茨公式, 就可得到
$$ {\int }_{0}^{1}\sqrt{1 - {x}^{2}}\mathrm{\;d}x = {\left. \left( \frac{1}{2}\arcsin x + \frac{x}{2}\sqrt{1 - {x}^{2}}\right) \right| }_{0}^{1} = \frac{\pi }{4}. $$
如果用换元法计算该积分,则可令 $x = \sin t$ ,于是
$$ {\int }_{0}^{1}\sqrt{1 - {x}^{2}}\mathrm{\;d}x = {\int }_{0}^{\pi /2}{\cos }^{2}t\mathrm{\;d}t $$
$$ = {\int }_{0}^{\pi /2}\frac{1 + \cos {2t}}{2}\mathrm{\;d}t $$
$$ = {\left. \frac{1}{2}\left( t + \frac{\sin {2t}}{2}\right) \right| }_{0}^{\pi /2} = \frac{\pi }{4}. $$
💡 答案与解析
解 如果求出 $\sqrt{1 - {x}^{2}}$ 的原函数
$$ \frac{1}{2}\arcsin x + \frac{x}{2}\sqrt{1 - {x}^{2}}, $$
再利用牛顿-莱布尼茨公式, 就可得到
$$ {\int }_{0}^{1}\sqrt{1 - {x}^{2}}\mathrm{\;d}x = {\left. \left( \frac{1}{2}\arcsin x + \frac{x}{2}\sqrt{1 - {x}^{2}}\right) \right| }_{0}^{1} = \frac{\pi }{4}. $$
如果用换元法计算该积分,则可令 $x = \sin t$ ,于是
$$ {\int }_{0}^{1}\sqrt{1 - {x}^{2}}\mathrm{\;d}x = {\int }_{0}^{\pi /2}{\cos }^{2}t\mathrm{\;d}t $$
$$ = {\int }_{0}^{\pi /2}\frac{1 + \cos {2t}}{2}\mathrm{\;d}t $$
$$ = {\left. \frac{1}{2}\left( t + \frac{\sin {2t}}{2}\right) \right| }_{0}^{\pi /2} = \frac{\pi }{4}. $$