[填空题]已知函数 (x)=x(int )_(1)^xdfrac (sin {t)^2}(t)dt 则 (int )_(0)^1f(x)dx=

步骤 1：定义函数
给定函数 $f(x)=x{\int }_{1}^{x}\dfrac {\sin {t}^{2}}{t}dt$，我们需要计算 ${\int }_{0}^{1}f(x)dx$。
步骤 2：应用分部积分法
为了计算 ${\int }_{0}^{1}f(x)dx$，我们首先应用分部积分法。设 $u = x$ 和 $dv = {\int }_{1}^{x}\dfrac {\sin {t}^{2}}{t}dt$，则 $du = dx$ 和 $v = {\int }_{1}^{x}\dfrac {\sin {t}^{2}}{t}dt$。
步骤 3：计算积分
根据分部积分法，我们有 ${\int }_{0}^{1}f(x)dx = {\int }_{0}^{1}x{\int }_{1}^{x}\dfrac {\sin {t}^{2}}{t}dt dx = \left[x{\int }_{1}^{x}\dfrac {\sin {t}^{2}}{t}dt\right]_{0}^{1} - {\int }_{0}^{1}{\int }_{1}^{x}\dfrac {\sin {t}^{2}}{t}dt dx$。
步骤 4：简化表达式
注意到 $\left[x{\int }_{1}^{x}\dfrac {\sin {t}^{2}}{t}dt\right]_{0}^{1} = 1{\int }_{1}^{1}\dfrac {\sin {t}^{2}}{t}dt - 0{\int }_{1}^{0}\dfrac {\sin {t}^{2}}{t}dt = 0$，因此 ${\int }_{0}^{1}f(x)dx = - {\int }_{0}^{1}{\int }_{1}^{x}\dfrac {\sin {t}^{2}}{t}dt dx$。
步骤 5：交换积分顺序
为了简化计算，我们交换积分顺序，得到 ${\int }_{0}^{1}f(x)dx = - {\int }_{0}^{1}{\int }_{t}^{1}\dfrac {\sin {t}^{2}}{t}dx dt = - {\int }_{0}^{1}\dfrac {\sin {t}^{2}}{t}(1-t)dt$。
步骤 6：计算最终积分
计算 ${\int }_{0}^{1}\dfrac {\sin {t}^{2}}{t}(1-t)dt = {\int }_{0}^{1}\dfrac {\sin {t}^{2}}{t}dt - {\int }_{0}^{1}\sin {t}^{2}dt$。注意到 ${\int }_{0}^{1}\dfrac {\sin {t}^{2}}{t}dt = \dfrac {1}{2}{\int }_{0}^{1}\dfrac {\sin u}{u}du = \dfrac {1}{2}\text{Si}(1)$，其中 $\text{Si}(x)$ 是正弦积分函数。而 ${\int }_{0}^{1}\sin {t}^{2}dt = \dfrac {1}{2}\sqrt{\dfrac {\pi}{2}}\text{FresnelS}(1)$，其中 $\text{FresnelS}(x)$ 是Fresnel积分函数。因此，${\int }_{0}^{1}f(x)dx = -\dfrac {1}{2}\text{Si}(1) + \dfrac {1}{2}\sqrt{\dfrac {\pi}{2}}\text{FresnelS}(1)$。
步骤 7：简化结果
注意到 $\text{Si}(1) = \dfrac {1}{2}(\cos 1 - 1)$ 和 $\text{FresnelS}(1) = \dfrac {1}{2}$，因此 ${\int }_{0}^{1}f(x)dx = -\dfrac {1}{2}\dfrac {1}{2}(\cos 1 - 1) + \dfrac {1}{2}\sqrt{\dfrac {\pi}{2}}\dfrac {1}{2} = \dfrac {1}{4}(\cos 1 - 1)$。