题目
计算下列定积分(1)(int )_(-1)^1dfrac (x)(sqrt {5-4x)}dx
计算下列定积分
(1)
题目解答
答案
令
则
∴原式
综上所述
解析
步骤 1：换元
令 $t = \sqrt{5-4x}$，则 $x = \dfrac{5-t^2}{4}$，$dx = -\dfrac{t}{2}dt$。
步骤 2：代入
将 $x$ 和 $dx$ 的表达式代入原积分，得到
${\int }_{-1}^{1}\dfrac {x}{\sqrt {5-4x}}dx = {\int }_{3}^{1}\dfrac {5-{t}^{2}}{4t}\cdot (-\dfrac {t}{2})dt$。
步骤 3：化简
化简得到
${\int }_{1}^{3}\dfrac {5-{t}^{2}}{4t}\cdot \dfrac {t}{2}dt = {\int }_{1}^{3}\dfrac {5-{t}^{2}}{8}dt$。
步骤 4：积分
计算积分
${\int }_{1}^{3}\dfrac {5-{t}^{2}}{8}dt = \dfrac {1}{8}{\int }_{1}^{3}(5-{t}^{2})dt$。
步骤 5：求解
求解得到
$\dfrac {1}{8}{\int }_{1}^{3}(5-{t}^{2})dt = \dfrac {1}{8}[\dfrac{5t}{1}-\dfrac{t^3}{3}]_{1}^{3} = \dfrac {1}{8}[(15-9)-(5-1/3)] = \dfrac {1}{8}[(6)-(14/3)] = \dfrac {1}{8}[(18-14)/3] = \dfrac {1}{8}[(4)/3] = \dfrac {1}{6}$。
令 $t = \sqrt{5-4x}$，则 $x = \dfrac{5-t^2}{4}$，$dx = -\dfrac{t}{2}dt$。
步骤 2：代入
将 $x$ 和 $dx$ 的表达式代入原积分，得到
${\int }_{-1}^{1}\dfrac {x}{\sqrt {5-4x}}dx = {\int }_{3}^{1}\dfrac {5-{t}^{2}}{4t}\cdot (-\dfrac {t}{2})dt$。
步骤 3：化简
化简得到
${\int }_{1}^{3}\dfrac {5-{t}^{2}}{4t}\cdot \dfrac {t}{2}dt = {\int }_{1}^{3}\dfrac {5-{t}^{2}}{8}dt$。
步骤 4：积分
计算积分
${\int }_{1}^{3}\dfrac {5-{t}^{2}}{8}dt = \dfrac {1}{8}{\int }_{1}^{3}(5-{t}^{2})dt$。
步骤 5：求解
求解得到
$\dfrac {1}{8}{\int }_{1}^{3}(5-{t}^{2})dt = \dfrac {1}{8}[\dfrac{5t}{1}-\dfrac{t^3}{3}]_{1}^{3} = \dfrac {1}{8}[(15-9)-(5-1/3)] = \dfrac {1}{8}[(6)-(14/3)] = \dfrac {1}{8}[(18-14)/3] = \dfrac {1}{8}[(4)/3] = \dfrac {1}{6}$。