ข้อสอบ PAT 1 - มีนาคม 2559

$\mathrm{พิจารณา} \left|x+2\right| = \left\{\begin{array}{ll}-\left(x+2\right) ;& \mathrm{โดย} x+2<0 \to x<-2\\ x+2 ;& \mathrm{โดย} x+2\ge 0 \to x\ge -2\end{array}\right.$

$$\begin{gathered} \mathrm{จากโจทย์} {\int }_{-4}^{-2}\frac{x^3+x^2+x}{x\left|x+2\right|-x^2-2}dx \\ \mathrm{ช่วงที่อินทิเกรตคือ} \left[-4,-2\right] \mathrm{จะได้} \left|x+2\right|=-\left(x+2\right) \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} {\int }_{-4}^{-2}\frac{x^3+x^2+x}{x\left|x+2\right|-x^2-2}dx \\ = {\int }_{-4}^{-2}\frac{x^3+x^2+x}{x(-x-2)-x^2-2}dx \\ = {\int }_{-4}^{-2}\frac{x^3+x^2+x}{-2x^2-2x-2}dx \\ = {\int }_{-4}^{-2}\frac{x\left(x^2+x+1\right)}{-2\left(x^2+x+1\right)}dx \end{gathered}$$

$$\begin{gathered} = {\int }_{-4}^{-2}\frac{x}{-2}dx = -\frac{1}{2}{\overline{)\left(\frac{x^2}{2}\right)}}_{-4}^{-2} \\ = -\frac{1}{2}\overline{)\left[\frac{{\left(-2\right)}^2}{2}-\frac{{\left(-4\right)}^2}{2}\right]} \\ = -\frac{1}{2}\left(2-8\right) \\ = 3 \end{gathered}$$

$\mathrm{ค่าของ} {\int }_{-4}^{-2}\frac{x^3+x^2+x}{x\left|x+2\right|-x^2-2}dx = \mathbf{3}$