جد المساحة المحددة بالدالة $y=f\left(x\right)=x^3-3x^2+2x$ ومحور السينات؟

$$\begin{gathered} \begin{array}{rl}& x^3-3x^2+2x=0\\ & x(x^2-3x+2)=0\mathrm{إما} x=0\\ & (x-2)(x-1)=0\Rightarrow \mathrm{إما} x=2\text{, أو }x=1\end{array} \\ \begin{array}{rl}& A=\left|{\int }_0^1(x^3-3x^2+2x)dx\right|+\left|{\int }_1^2(x^3-3x^2+2x)dx\right|=\left|{[\frac{x^4}{4}-x^3+x^2]}_0^1\right|+\left|{[\frac{x^4}{4}-x^3+x^2]}_0^1\right|\\ & A=|(\frac{1}{4}-1+1)-(0\left)\right|+\left|\right(4-8+4)-(\frac{1}{4}-1+1)|\\ & A=\left|\frac{1}{4}\right|+\left|\frac{-1}{4}\right|==\frac{1}{4}+\frac{1}{4}=\frac{1}{2} {\text{unit}}^2\end{array} \end{gathered}$$

تمارين (4-4)

(1)- جد المساحة بين منحني الدالة (f(x ومحور السينات والمستقيمين 2=x=-$y=f\left(x\right)=x^3-4x$؟

$$\begin{gathered} \begin{array}{rl}& x^3-4x=0\Rightarrow x(x^2-4)=0\\ & \mathrm{إما} x=0\in [-2,2],\mathrm{أو} (x-2)(x+2)=0\end{array} \\ \begin{array}{rl}& x=2\in [-2,1]\\ & x=-2\in [-2.1]\\ & [-2,0],[0,2]\end{array} \\ \begin{array}{rl}& A=\left|{\int }_{-2}^0(x^3-4x)dx\right|+\left|{\int }_0^2(x^2-4x)dx\right|=\left|{[\frac{x^4}{4}-2x^2]}_{-2}^0\right|+\left|{[\frac{x^4}{4}-2x^2]}_0^2\right|\\ & A=\left|\right(0)-(\frac{16}{4}-8)|+|(\frac{16}{4}-8)-(0\left)\right|\\ & A=|-4+8|+\left|\right(4-8\left)\right|=\left|4\right|+|-4|\\ & A=4+4=8 {\text{unit}}^2\end{array} \end{gathered}$$

(2)- جد المسافة المحددة بمنحني الدالة $y=f\left(x\right)=x^4-x^2$ ومحور السينات وعلى الفترة [1,1-]؟

$$\begin{gathered} x^4-x^2=0\Rightarrow x^2(x^2-1)=0 \\ \mathrm{إما} x^2=0\Rightarrow x=0\in [-1.1],\mathrm{أو} (x-1)(x+1)=0 \\ \begin{array}{rl}& x=-1\in [-1,1]\\ & x=-1\in [-1,1]\\ & [-1,0],[0,1]\end{array} \\ \begin{array}{rl}& A=\left|{\int }_{-1}^0(x^4-x^2)dx\right|+\left|{\int }_0^1(x^4-x^2)dx\right|\Rightarrow A=\left|{[\frac{x^5}{5}-\frac{x^3}{3}]}_{-1}^0\right|+\left|{[\frac{x^5}{5}-\frac{x^3}{3}]}_0^1\right|\\ & A=\left|\right(0)-(-\frac{1}{5}+\frac{1}{3})|+|(\frac{1}{5}-\frac{1}{3})-(0\left)\right|\Rightarrow A=\left|\frac{3-5}{15}\right|+\left|\frac{3-5}{15}\right|=\left|\frac{-2}{15}\right|+\left|\frac{-2}{4}\right|\\ & A=\frac{2}{15}+\frac{2}{15}=\frac{4}{15} {\text{unit}}^2\end{array} \end{gathered}$$

(3)- جد المساحة المحددة بالدالة $y=f\left(x\right)=x^3-3x^2+2x$ ومحور السينات؟

$$\begin{gathered} \begin{array}{rl}& x^3-3x^2+2x=0\\ & x(x^2-3x+2)=0\mathrm{إما} x=0\\ & (x-2)(x-1)=0\Rightarrow \mathrm{إما} x=2\text{, أو }x=1\end{array} \\ \begin{array}{rl}& A=\left|{\int }_0^1(x^3-3x^2+2x)dx\right|+\left|{\int }_1^2(x^3-3x^2+2x)dx\right|=\left|{[\frac{x^4}{4}-x^3+x^2]}_0^1\right|+\left|{[\frac{x^4}{4}-x^3+x^2]}_0^1\right|\\ & A=|(\frac{1}{4}-1+1)-(0\left)\right|+\left|\right(4-8+4)-(\frac{1}{4}-1+1)|\\ & A=\left|\frac{1}{4}\right|+\left|\frac{-1}{4}\right|==\frac{1}{4}+\frac{1}{4}=\frac{1}{2} {\text{unit}}^2\end{array} \end{gathered}$$

(4)- جد المساحة المحددة بمنحني الدالتين$g\left(x\right)=\frac{1}{2}x,f\left(x\right)\sqrt{x-1}$ والمستقيمين x=2,x=5؟

$$\begin{gathered} f\left(x\right)=g\left(x\right)\Rightarrow \sqrt{x-1}=\frac{1}{2}x \mathrm{الطرفين} \mathrm{بتربيع} \\ \begin{array}{rl}& x-1=\frac{1}{4}{x}^2\Rightarrow [\frac{1}{4}{x}^2-x+1=0]\left(4\right)\\ & x^2-4x+4=0\Rightarrow (x-2{)}^2=0 \mathrm{التربيعي} \mathrm{بالجذر}\Rightarrow x=2\in [2,5]\\ & A=|{\int }_2^5\frac{1}{2}x-\sqrt{x-1}dx|=|{\int }_2^5\frac{1}{2}x-(x+1{)}^{\frac{3}{2}}dx|\\ & A=\left|{[\frac{x^2}{4}-\frac{2}{3}(x-1{)}^{\frac{3}{2}}]}_2^5\right|=\mid [(\frac{25}{4}-\frac{2}{3}(2{)}^3)-{(1-\frac{2}{3}(1{)}^3]}_2^5\mid \\ & A=|\frac{75-64}{12}-\frac{1}{3}|=|\frac{11}{12}-\frac{1}{3}|=\frac{11-4}{12}=\frac{7}{12} uni{t}^3\end{array} \end{gathered}$$

(5)- جد المساحة المحددة بمنحني الدالتين$y=x^2,y=x^4-12$

$\begin{array}{rl}& x^4-x^2-12=0\Rightarrow (x^2+4)(x^2+3)=0\\ & x=\mp 2[-2,2]\\ & {A=\left|{\int }_{-2}^2(x^4-x^2+12)dx\right|=\left|\right|\frac{x^5}{5}-\frac{x^3}{3}+12x]}_{-2}^2\mid \\ & A=\left|{[\frac{32}{5}-\frac{8}{3}-24]}_0^1\right|-\left|{[\frac{32}{5}+\frac{8}{3}+24]}_0^1\right|\\ & A=|\frac{32}{5}-\frac{8}{3}-24+\frac{32}{5}-\frac{8}{3}-24|=|\frac{64}{5}-\frac{16}{3}-48|\\ & A=\left|\frac{192-80-720}{15}\right|=\left|\frac{-608}{15}\right|=\frac{608}{15} {\text{unit}}^2\end{array}$