أتدرب وأحل المسائل

أتدرب وأحل المسائل

التكامل بالتعويض

أجد كلاً من التكاملات الآتية:

$\int \frac{x}{\sqrt{x^{2} + 4}} d x$ (1)

$\begin{aligned} \int \frac{x}{\sqrt{x^{2} + 4}} d x \\ u = x^{2} + 4 & \Rightarrow \frac{d u}{d x} = 2 x \\ \Rightarrow d x & = \frac{d u}{2 x} \\ \int \frac{x}{\sqrt{x^{2} + 4}} d x & = \int \frac{x}{\sqrt{u}} \times \frac{d u}{2 x} \\ = \int \frac{1}{2 \sqrt{u}} d u \\ = \int \frac{1}{2} u^{- \frac{1}{2}} d u \\ = u^{\frac{1}{2}} + C \\ = \sqrt{x^{2} + 4} + C \end{aligned}$

$\int x^{2} {(2 x^{3} + 5)}^{4} d x$ (2)

$\begin{aligned} \int x^{2} {(2 x^{3} + 5)}^{4} d x \\ u = 2 x^{3} + 5 \Rightarrow \frac{d u}{d x} = 6 x^{2} \\ \Rightarrow d x = \frac{d u}{6 x^{2}} \\ \int x^{2} {(2 x^{3} + 5)}^{4} d x = \int x^{2} u^{4} \times \frac{d u}{6 x^{2}} \\ = \int \frac{1}{6} u^{4} d u \\ = \frac{1}{30} u^{5} + C \\ = \frac{1}{30} {(2 x^{3} + 5)}^{5} + C \end{aligned}$

$\int 3 x \sqrt{x^{2} + 7} d x$ (3)

$\begin{aligned} \int 3 x \sqrt{x^{2} + 7} d x \\ \begin{aligned} u = x^{2} + 7 \Rightarrow \frac{d u}{d x} & = 2 x \\ \Rightarrow d x & = \frac{d u}{2 x} \\ \int 3 x \sqrt{x^{2} + 7} d x & = \int 3 x \sqrt{u} \times \frac{d u}{2 x} \\ = \int \frac{3}{2} u^{\frac{1}{2}} d u \\ = u^{\frac{3}{2}} + C \\ = \sqrt{{(x^{2} + 7)}^{3}} + C \end{aligned} \end{aligned}$

$\int x^{6} e^{1 - x^{7}} d x$ (4)

$\begin{aligned} \int x^{6} e^{1 - x^{7}} d x \\ u = 1 - x^{7} \Rightarrow \frac{d u}{d x} = - 7 x^{6} \\ \Rightarrow d x = \frac{d u}{- 7 x^{6}} \\ \int x^{6} e^{1 - x^{7}} d x = \int x^{6} e^{u} \times \frac{d u}{- 7 x^{6}} \\ = \int - \frac{1}{7} e^{u} d u \\ = - \frac{1}{7} e^{u} + C \\ = - \frac{1}{7} e^{1 - x^{7}} + C \end{aligned}$

$\int \frac{x^{4}}{{(x^{5} + 9)}^{3}} d x$ (5)

$\begin{aligned} \int \frac{x^{4}}{{(x^{5} + 9)}^{3}} d x \\ u = x^{5} + 9 \Rightarrow \frac{d u}{d x} = 5 x^{4} \\ \Rightarrow d x = \frac{d u}{5 x^{4}} \\ \int \frac{x^{4}}{{(x^{5} + 9)}^{3}} d x = \int \frac{x^{4}}{u^{3}} \times \frac{d u}{5 x^{4}} \\ = \int \frac{1}{5} u^{- 3} d u \\ = - \frac{1}{10} u^{- 2} + C \\ = - \frac{1}{10 {(x^{5} + 9)}^{2}} + C \end{aligned}$

$\int (3 x^{2} - 1) e^{x^{3} - x} d x$ (6)

$\begin{aligned} \int (3 x^{2} - 1) e^{x^{3} - x} d x \\ u = x^{3} - x \Rightarrow \frac{d u}{d x} = 3 x^{2} - 1 \\ \Rightarrow d x = \frac{d u}{3 x^{2} - 1} \\ \int (3 x^{2} - 1) e^{x^{3} - x} d x = \int (3 x^{2} - 1) e^{u} \frac{d u}{3 x^{2} - 1} \\ = \int e^{u} d u \\ = e^{u} + C \\ = e^{x^{3} - x} + C \end{aligned}$

$\int \frac{3 x - 3}{\sqrt{x^{2} - 2 x + 4}} d x$ (7)

$b b b \begin{aligned} \int \frac{3 x - 3}{\sqrt{x^{2} - 2 x + 4}} d x \\ u = x^{2} - 2 x + 4 & \Rightarrow \frac{d u}{d x} = 2 x - 2 \\ \Rightarrow d x = \frac{d u}{2 x - 2} \\ \int \frac{3 x - 3}{\sqrt{x^{2} - 2 x + 4}} d x & = \int \frac{3 x - 3}{\sqrt{u}} \times \frac{d u}{2 x - 2} \\ = \int \frac{3 (x - 1)}{\sqrt{u}} \times \frac{d u}{2 (x - 1)} \\ = \int \frac{3}{2} u^{- \frac{1}{2}} d u \\ = 3 u^{\frac{1}{2}} + C \\ = 3 \sqrt{x^{2} - 2 x + 4} + C \end{aligned}$

$\int \frac{1}{x \ln x} d x$ (8)

$\begin{aligned} \int \frac{1}{x \ln x} d x \\ u = \ln x \Rightarrow \frac{d u}{d x} = \frac{1}{x} \\ \Rightarrow d x = x d u \\ \int \frac{1}{x \ln x} d x = \int \frac{1}{x u} \times x d u \\ = \int \frac{1}{u} d u \\ = \ln | u | + C \\ = \ln | \ln x | + C \end{aligned}$

$\int \sin x (1 + \cos x)^{4} d x$ (9)

$\begin{aligned} \int \sin x (1 + \cos x)^{4} d x \\ u = 1 + \cos x \Rightarrow \frac{d u}{d x} = - \sin x \\ \Rightarrow d x = \frac{d u}{- \sin x} \\ \int \sin x (1 + \cos x)^{4} d x = \int \sin x u^{4} \times \frac{d u}{- \sin x} \\ = \int - u^{4} d u \\ = - \frac{1}{5} u^{5} + C \\ = - \frac{1}{5} (1 + \cos x)^{5} + C \end{aligned}$

$\int \sin^{5} 2 x \cos 2 x d x$ (10)

$b b b \begin{aligned} \int \sin^{5} 2 x \cos 2 x d x \\ u = \sin 2 x \Rightarrow \frac{d u}{d x} = 2 \cos 2 x \\ \Rightarrow d x = \frac{d u}{2 \cos 2 x} \\ \int \sin^{5} 2 x \cos 2 x d x = \int u^{5} \cos 2 x \times \frac{d u}{2 \cos 2 x} \\ = \int \frac{1}{2} u^{5} d u \\ = \frac{1}{12} u^{6} + C \\ = \frac{1}{12} (\sin 2 x)^{6} + C \end{aligned}$

$\int \frac{\sin (\frac{1}{x})}{x^{2}} d x$ (11)

$\begin{aligned} \int \frac{\sin (\frac{1}{x})}{x^{2}} d x \\ u = \frac{1}{x} \Rightarrow \frac{d u}{d x} = - \frac{1}{x^{2}} \\ \Rightarrow d x = - x^{2} d u \\ \int \frac{\sin (\frac{1}{x})}{x^{2}} d x = \int \frac{\sin (u)}{x^{2}} \times - x^{2} d u \\ = \int - \sin u d u \\ = \cos u + C \\ = \cos (\frac{1}{x}) + C \end{aligned}$

$\int \frac{\cos x}{e^{\sin x}} d x$ (12)

$\begin{aligned} \int \frac{\cos x}{e^{\sin x}} d x \\ u = \sin x \Rightarrow \frac{d u}{d x} = \cos x \\ \Rightarrow d x = \frac{d u}{\cos x} \\ \int \frac{\cos x}{e^{\sin x}} d x = \int \frac{\cos x}{e^{u}} \times \frac{d u}{\cos x} \\ = \int \frac{1}{e^{u}} d u \\ = \int e^{- u} d u \\ = - e^{- u} + C \\ = - e^{- \sin x} + C \\ = - \frac{1}{e^{\sin x}} + C \end{aligned}$

$\int e^{x} {(2 + e^{x})}^{5} d x$ (13)

$\begin{aligned} \int e^{x} {(2 + e^{x})}^{5} d x \\ u = 2 + e^{x} \Rightarrow \frac{d u}{d x} = e^{x} \\ \Rightarrow d x = \frac{d u}{e^{x}} \\ \int e^{x} {(2 + e^{x})}^{5} d x = \int e^{x} u^{5} \times \frac{d u}{e^{x}} \\ = \int u^{5} d u \\ = \frac{1}{6} u^{6} + C \\ = \frac{1}{6} {(2 + e^{x})}^{6} + C \end{aligned}$

$\int \frac{\cos (\ln x)}{x} d x$ (14)

$\begin{aligned} \int \frac{\cos (\ln x)}{x} d x \\ u = \ln x \Rightarrow \frac{d u}{d x} = \frac{1}{x} \\ \Rightarrow d x = x d u \\ \int \frac{\cos (\ln x)}{x} d x = \int \frac{\cos (u)}{x} \times x d u \\ = \int \cos u d u \\ = \sin u + C \\ = \sin (\ln x) + C \end{aligned}$

$\int (3 x^{2} - 2 x - 1) {(x^{3} - x^{2} - x)}^{4} d x$ (15)

$\begin{aligned} \begin{matrix} \int (3 x^{2} - 2 x - 1) {(x^{3} - x^{2} - x)}^{4} d x \\ u = x^{3} - x^{2} - x \Rightarrow \frac{d u}{d x} = 3 x^{2} - 2 x - 1 \end{matrix} \\ \Rightarrow d x = \frac{d u}{3 x^{2} - 2 x - 1} \\ \int (3 x^{2} - 2 x - 1) {(x^{3} - x^{2} - x)}^{4} d x \\ = \int (3 x^{2} - 2 x - 1) u^{4} \times \frac{d u}{3 x^{2} - 2 x - 1} \\ = \int u^{4} d u \\ = \frac{1}{5} u^{5} + C \\ = \frac{1}{5} {(x^{3} - x^{2} - x)}^{5} + C \end{aligned}$

أجد قيمة كل من التكاملات الآتية:

$\int_{0}^{2} (2 x - 1) e^{x^{2} - x} d x$ (16)

$\begin{aligned} \int_{0}^{2} (2 x - 1) e^{x^{2} - x} d x \\ u = x^{2} - x \Rightarrow \frac{d u}{d x} = 2 x - 1 \\ \Rightarrow d x = \frac{d u}{2 x - 1} \\ \begin{aligned} x = 2 ⟹ u = (2)^{2} - 2 = 2 \\ x = 0 \Rightarrow u = (0)^{2} - 0 = 0 \end{aligned} \\ \begin{aligned} \int_{0}^{2} (2 x - 1) e^{x^{2} - x} d x & = \int_{0}^{2} (2 x - 1) e^{u} \frac{d u}{2 x - 1} \\ = \int_{0}^{2} e^{u} d u \\ = {e^{u} |}_{0}^{2} \\ = e^{2} - e^{0} \\ = e^{2} - 1 \end{aligned} \end{aligned}$

$\int_{1}^{2} \frac{e^{1 / x}}{x^{2}} d x$ (17)

$\begin{aligned} \int_{1}^{2} \frac{e^{\frac{1}{x}}}{x^{2}} d x \\ u = \frac{1}{x} \Rightarrow \frac{d u}{d x} = - \frac{1}{x^{2}} \\ \Rightarrow d x = - x^{2} d u \\ x = 2 \Rightarrow u = \frac{1}{2} \\ x^{2} = \frac{1}{x^{\frac{1}{x}}} d x = \int_{1}^{\frac{1}{2}} \frac{e^{u}}{x^{2}} \times - x^{2} d u \\ = \int_{1}^{\frac{1}{2}} - e^{u} d u \\ = - {e^{u} |}_{1}^{\frac{1}{2}} \\ = - e^{\frac{1}{2}} + e \\ = - \sqrt{e} + e \end{aligned}$

$\int_{e}^{e^{3}} \frac{\sqrt{\ln x}}{x} d x$ (18)

$\begin{aligned} \int_{e}^{e^{3}} \frac{\sqrt{\ln x}}{x} d x \\ u = \ln x \Rightarrow \frac{d u}{d x} = \frac{1}{x} \\ \Rightarrow d x = x d u \\ x = e^{3} \Rightarrow u = \ln e^{3} = 3 \\ x = e \Rightarrow u = \ln e = 1 \\ \int_{e}^{e^{3}} \frac{\sqrt{\ln x}}{x} d x = \int_{1}^{3} \frac{\sqrt{u}}{x} x d u \\ = \int_{1}^{3} u^{\frac{1}{2}} d u \\ = {\frac{2}{3} u^{\frac{3}{2}} |}_{1}^{3} \\ = {\frac{2}{3} \sqrt{u^{3}} |}_{1}^{3} \\ = \frac{2}{3} \sqrt{3^{3}} - \frac{2}{3} \sqrt{1^{3}} \\ = 2 \sqrt{3} - \frac{2}{3} \end{aligned}$

$\int_{0}^{1} (x^{3} + x) \sqrt{x^{4} + 2 x^{2} + 1} d x$ (19)

$\begin{aligned} \int_{0}^{1} (x^{3} + x) \sqrt{x^{4} + 2 x^{2} + 1} d x \\ u = x^{4} + 2 x^{2} + 1 \Rightarrow \frac{d u}{d x} = 4 x^{3} + 4 x \\ \Rightarrow d x = \frac{d u}{4 x^{3} + 4 x} \\ \begin{aligned} x = 1 \Rightarrow u = (1)^{4} + 2 (1)^{2} + 1 & = 4 \\ x = 0 \Rightarrow u = (0)^{4} + 2 (0)^{2} & + 1 = 1 \\ \int_{0}^{1} (x^{3} + x) \sqrt{x^{4} + 2 x^{2} + 1} d x & = \int_{1}^{4} (x^{3} + x) \sqrt{u} \times \frac{d u}{4 x^{3} + 4 x} \\ = \int_{1}^{4} (x^{3} + x) \sqrt{u} \times \frac{d u}{4 (x^{3} + x)} \\ = \int_{1}^{4} \frac{1}{4} u^{\frac{1}{2}} d u \\ = {\frac{1}{6} u^{\frac{3}{2}} |}_{1}^{4} \\ = {\frac{1}{6} \sqrt{u^{3}} |}_{1}^{4} \\ = \frac{1}{6} \sqrt{4^{3}} - \frac{1}{6} \sqrt{1^{3}} \\ = \frac{7}{6} \end{aligned} \end{aligned}$

$\int_{0}^{3} \frac{x}{\sqrt{x^{2} + 1}} d x$ (20)

$\begin{aligned} \int_{0}^{3} \frac{x}{\sqrt{x^{2} + 1}} d x \\ u = x^{2} + 1 \Rightarrow \frac{d u}{d x} = 2 x \\ \Rightarrow d x = \frac{d u}{2 x} \\ x = 3 \Rightarrow u = 10 \\ x = 0 \Rightarrow u = 1 \\ \int_{0}^{3} \frac{x}{\sqrt{x^{2} + 1}} d x = \int_{1}^{10} \frac{x}{\sqrt{u}} \times \frac{d u}{2 x} \\ = \int_{1}^{10} \frac{1}{2} u^{- \frac{1}{2}} d u \\ = {u^{\frac{1}{2}} |}_{1}^{10} \\ = {\sqrt{u} |}_{1}^{10} \\ = \sqrt{10} - 1 \end{aligned}$

$\int_{1}^{2} \frac{2 x + 1}{{(x^{2} + x + 4)}^{3}} d x$ (21)

$\begin{aligned} \int_{1}^{2} \frac{2 x + 1}{{(x^{2} + x + 4)}^{3}} d x \\ u = x^{2} + x + 4 \Rightarrow \frac{d u}{d x} = 2 x + 1 \\ \Rightarrow d x = \frac{d u}{2 x + 1} \\ x = 2 \Rightarrow u = (2)^{2} + 2 + 4 = 10 \\ x = 1 \Rightarrow u = (1)^{2} + 1 + 4 = 6 \\ \int_{1}^{2} \frac{2 x + 1}{{(x^{2} + x + 4)}^{3}} d x = \int_{6}^{10} \frac{2 x + 1}{u^{3}} \times \frac{d u}{2 x + 1} \\ = \int_{6}^{10} u^{- 3} d u \\ = - {\frac{1}{2} u^{- 2} |}_{6}^{10} \\ = - {\frac{1}{2 u^{2}} |}_{6}^{10} \end{aligned}$

أجد مساحة المنطقة المظللة في كل من التمثيلين البيانيين الآتيين:

التمثيل البياني للسؤال 23

$A = - \int_{- 1}^{0} 6 x (x^{2} + 1) d x + \int_{0}^{1} 6 x (x^{2} + 1) d x$

هناك طريقتان للحل: إما التكامل بالتعويض، أو تكامل كثير حدود بعد توزيع الأقواس.

طريقة التكامل بالتعويض:

$\begin{aligned} u & = x^{2} + 1 \Rightarrow \frac{d u}{d x} = 2 x \\ \Rightarrow & \Rightarrow d x = \frac{d u}{2 x} \\ x & = 0 \Rightarrow u = (0)^{2} + 1 = 1 \\ x & = - 1 \Rightarrow u = (- 1)^{2} + 1 = 2 \\ x & = 1 \Rightarrow u = (1)^{2} + 1 = 2 \\ A & = - \int_{- 1}^{0} 6 x (x^{2} + 1) d x + \int_{0}^{1} 6 x (x^{2} + 1) d x \\ = - \int_{2}^{1} 6 x u \times \frac{d u}{2 x} + \int_{1}^{2} 6 x u \times \frac{d u}{2 x} \\ = - \int_{2}^{1} 3 u d u + \int_{1}^{2} 3 u d u \\ = - {\frac{3}{2} u^{2} |}_{2}^{1} + {\frac{3}{2} u^{2} |}_{1}^{2} \\ = - \frac{3}{2} (1)^{2} + \frac{3}{2} (2)^{2} + \frac{3}{2} (2)^{2} - \frac{3}{2} (1)^{2} \\ = 9 \end{aligned}$

ومنه مساحة المنطقة المظللة هي 9 وحدات مربعة.

التمثيل البياني للسؤال 24

$\begin{aligned} A = - \int_{- 4}^{0} x \sqrt{16 - x^{2}} d x + \int_{0}^{4} x \sqrt{16 - x^{2}} d x \\ u = 16 - x^{2} \Rightarrow \frac{d u}{d x} = - 2 x \\ \Rightarrow d x = \frac{d u}{- 2 x} \\ x = 0 \Rightarrow u = 16 - (0)^{2} = 16 \\ x = - 4 \Rightarrow u = 16 - (- 4)^{2} = 0 \\ x = 4 \Rightarrow u = 16 - (4)^{2} = 0 \\ A = - \int_{- 4}^{0} x \sqrt{16 - x^{2}} d x + \int_{0}^{4} x \sqrt{16 - x^{2}} d x \\ = - \int_{0}^{16} x \sqrt{u} \times \frac{d u}{- 2 x} + \int_{16}^{0} x \sqrt{u} \times \frac{d u}{- 2 x} \\ = \int_{0}^{16} \frac{1}{2} u^{\frac{1}{2}} d u + \int_{16}^{0} - \frac{1}{2} u^{\frac{1}{2}} d u \\ = {\frac{1}{3} u^{\frac{3}{2}} |}_{0}^{16} + - {\frac{1}{3} u^{\frac{3}{2}} |}_{16}^{0} \\ = {\frac{1}{3} \sqrt{u^{3}} |}_{0}^{16} + - {\frac{1}{3} \sqrt{u^{3}} |}_{16}^{0} \\ = \frac{1}{3} \sqrt{(16)^{3}} - \frac{1}{3} \sqrt{(0)^{3}} - \frac{1}{3} \sqrt{(0)^{3}} + \frac{1}{3} \sqrt{(16)^{3}} \\ = \frac{128}{3} \end{aligned}$

ومنه مساحة المنطقة المظللة هي $\frac{128}{3}$ وحدات مربعة.

في كل مما يأتي المشتقة الأولى للاقتران $f (x)$ ، ونقطة يمر بها منحنى $y = f (x)$ أستعمل المعلومات المعطاة لإيجاد قاعدة الاقتران $f (x)$ :

$f^{'} (x) = x e^{4 - x^{2}}; (- 2, 1)$ (24)

$\begin{aligned} f (x) & = \int x e^{4 - x^{2}} d x \\ u = & 4 - x^{2} \Rightarrow \frac{d u}{d x} = - 2 x \\ \Rightarrow d x = \frac{d u}{- 2 x} \\ f (x) & = \int x e^{4 - x^{2}} d x \\ = \int x e^{u} \frac{d u}{- 2 x} \\ = \int - \frac{1}{2} e^{u} d u \\ = - \frac{1}{2} e^{u} + C \\ = - \frac{1}{2} e^{4 - x^{2}} + C \end{aligned}$

لإيجاد ثابت التكامل، نعوض النقطة $(- 2, 1)$ :

$\begin{aligned} f (x) = - \frac{1}{2} e^{4 - x^{2}} + C \Rightarrow f (- 2) = - \frac{1}{2} e^{4 - (- 2)^{2}} + C \\ \Rightarrow 1 = - \frac{1}{2} + C \\ \Rightarrow C = \frac{3}{2} \\ f (x) = - \frac{1}{2} e^{4 - x^{2}} + \frac{3}{2} \end{aligned}$

$f^{'} (x) = \frac{2 x}{{(1 - x^{2})}^{2}}; (0, - 1)$ (25)

$\begin{aligned} f (x) & = \int \frac{2 x}{{(1 - x^{2})}^{2}} d x \\ u = 1 & \Rightarrow \frac{d u}{d x} = - 2 x \\ \Rightarrow d x = \frac{d u}{- 2 x} \\ f (x) & = \int \frac{2 x}{{(1 - x^{2})}^{2}} d x \\ = \int \frac{2 x}{u^{2}} \times \frac{d u}{- 2 x} \\ = \int - u^{- 2} d u \\ = u^{- 1} + C \\ = \frac{1}{1 - x^{2}} + C \end{aligned}$

لإيجاد ثابت التكامل، نعوض النقطة $(0, - 1)$ :

$\begin{aligned} f (x) = \frac{1}{1 - x^{2}} + C & \Rightarrow f (0) = \frac{1}{1 - 0^{2}} + C \\ \Rightarrow - 1 = 1 + C \\ \Rightarrow C = - 2 \\ f (x) = \frac{1}{1 - x^{2}} - 2 \end{aligned}$

(26) يتحرك جسيم في مسار مستقيم، وتعطى سرعته المتجهة بالاقتران: $ι (t) = \frac{- 2 t}{\sqrt{{(1 + t^{2})}^{3}}}$ ، حيث $t$ الزمن بالثواني، و $v$ سرعته المتجهة بالمتر لكل ثانية. إذا كان الموقع الابتدائي للجسيم $4 m$ ، فأجد موقع الجسيم بعد $t$ ثانية من بدء الحركة.

$\begin{aligned} s (t) = \int \frac{- 2 t}{\sqrt{{(1 + t^{2})}^{3}}} d t \\ u = 1 + t^{2} \Rightarrow \frac{d u}{d t} = 2 t \\ \Rightarrow d t = \frac{d u}{2 t} \\ \int \frac{- 2 t}{\sqrt{{(1 + t^{2})}^{3}}} d t = \int \frac{- 2 t}{\sqrt{u^{3}}} \times \frac{d u}{2 t} \\ = \int - u^{- \frac{3}{2}} d u \\ = 2 u^{- \frac{1}{2}} + C \\ = 2 {(1 + t^{2})}^{- \frac{1}{2}} + C \\ = \frac{2}{\sqrt{1 + t^{2}}} + C \end{aligned}$

بما أن الموقع الابتدائي للجسيم $4 m$ ، إذن، $s (0) = 4$ :

$\begin{aligned} s (t) = \frac{2}{\sqrt{1 + t^{2}}} + C \Rightarrow f (0) = \frac{2}{\sqrt{1 + 0^{2}}} + C \\ \Rightarrow 4 = 2 + C \\ \Rightarrow C = 2 \\ s (t) = \frac{2}{\sqrt{1 + t^{2}}} + 2 \end{aligned}$

أرض زراعية (27) زراعة: يمثل الاقتران $V (t)$ سعر دونم أرض زراعية في الأغوار الأردنية (بالدينار) بعد $t$ سنة من الآن. إذا كان: $V^{'} (t) = \frac{0.4 t^{3}}{\sqrt[3]{0.2 t^{4} + 8000}}$ هو معدل التغير في سعر دونم الأرض، فأجد $V (t)$ ، علماً بأن سعره الآن $JD 5000$ .

$\begin{aligned} V (t) = \int \frac{0.4 t^{3}}{\sqrt[3]{0.2 t^{4} + 8000}} d t \\ u = 0.2 t^{4} + 8000 \Rightarrow \frac{d u}{d t} = 0.8 t^{3} \\ \Rightarrow d t = \frac{d u}{0.8 t^{3}} \\ V (t) = \int \frac{0.4 t^{3}}{\sqrt[3]{0.2 t^{4} + 8000}} d x \\ = \int \frac{0.4 t^{3}}{\sqrt[3]{u}} \times \frac{d u}{0.8 t^{3}} \\ = \int \frac{1}{2} u^{- \frac{1}{3}} d u \\ = \frac{1}{3} u^{\frac{2}{3}} + C \\ = \frac{1}{3} \sqrt[3]{u^{2}} + C \\ = \frac{1}{3} \sqrt[3]{{(0.2 t^{4} + 8000)}^{2}} + C \end{aligned}$

بما أن سعر دونم الأرض الآن هو 5000 دينار، إذن، $V (0) = 5000$ ومنه:

$\begin{aligned} V (t) = \frac{1}{3} \sqrt[3]{{(0.2 t^{4} + 8000)}^{2}} + C \\ V (0) = \frac{1}{3} \sqrt[3]{{(0.2 (0)^{4} + 8000)}^{2}} + C \\ 5000 = \frac{1}{3} \sqrt[3]{(8000)^{2}} + C \\ 5000 = \frac{400}{3} + C \\ C = \frac{14600}{3} \\ V (t) = \frac{1}{3} \sqrt[3]{{(0.2 t^{4} + 8000)}^{2}} + \frac{14600}{3} \end{aligned}$

(28) سكان: أشارت دراسة إلى أن عدد السكان في إحدى المدن يتغير سنوياً بمعدل يمكن نمذجته بالاقتران: $P^{'} (t) = \frac{4 e^{0.2 t}}{\sqrt{4 + e^{0.2 t}}}$ ، حيث $t$ عدد السنوات منذ عام 2015 م، و $P (t)$ عدد السكان بالآلاف. أجد مقدار الزيادة في عدد السكان عام 2015 م إلى عام 2025 م.

$\begin{aligned} \Rightarrow d t = \frac{d u}{0.2 e^{0.2 t}} \\ t = 10 \Rightarrow u = 4 + e^{0.2 (10)} = 4 + e^{2} \\ t = 0 \Rightarrow u = 4 + e^{0.2 (0)} = 5 \\ \int_{0}^{10} \frac{4 e^{0.2 t}}{\sqrt{4 + e^{0.2 t}}} d t = \int_{5}^{4 + e^{2}} \frac{4 e^{0.2 t}}{\sqrt{u}} \times \frac{d u}{0.2 e^{0.2 t}} \\ = \int_{5}^{4 + e^{2}} 20 u^{- \frac{1}{2}} d u \\ = {40 u^{\frac{1}{2}} |}_{5}^{4 + e^{2}} \\ = {40 \sqrt{u} |}_{5}^{4 + e^{2}} \\ = 40 \sqrt{4 + e^{2}} - 40 \sqrt{5} \\ \approx 46 \end{aligned}$

إذن يزداد عدد سكان هذه المدينة بحوالي 46 ألف شخص من 2015 م إلى 2025 م.

إعداد : شبكة منهاجي التعليمية

11 / 02 / 2023

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