إجابات كتاب التمارين

إجابات كتاب التمارين

التكامل بالتعويض

أجد كلاً من التكاملات الآتية:

$\int x \sqrt{x^{2} + 3} d x$ (1)

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

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

$\begin{aligned} \int x^{4} e^{x^{5} + 2} d x \\ \begin{aligned} u = x^{5} + 2 \Rightarrow \frac{d u}{d x} = 5 x^{4} \Rightarrow d x = \frac{d u}{5 x^{4}} \\ \int x^{4} e^{x^{5} + 2} d x = \int x^{4} e^{u} \frac{d u}{5 x^{4}} = \int_{0} \frac{1}{5} e^{u} d u \\ = \frac{1}{5} e^{u} + C = \frac{1}{5} e^{x^{5} + 2} + C \end{aligned} \end{aligned}$

$\int (x + 1) {(x^{2} + 2 x + 5)}^{4} d x$ (3)

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

$\int \frac{(\ln x)^{3}}{x} d x$ (4)

$\begin{aligned} \int \frac{(\ln x)^{3}}{x} d x \\ u = \ln x \Rightarrow \frac{d u}{d x} = \frac{1}{x} \Rightarrow d x = x d u \\ \begin{aligned} \int \frac{(\ln x)^{3}}{x} d x = \int \frac{u^{3}}{x} x d u & = \int_{0} u^{3} d u \\ = \frac{1}{4} u^{4} + C = \frac{1}{4} (\ln x)^{4} + C \end{aligned} \end{aligned}$

$\int \frac{\cos x}{\sin^{4} x} d x$ (5)

$\begin{aligned} \int \frac{\cos x}{\sin^{4} x} d x \\ u = \sin x \Rightarrow \frac{d u}{d x} = \cos x \Rightarrow d x = \frac{d u}{\cos x} \\ \int \frac{\cos x}{\sin^{4} x} d x = \int \frac{\cos x}{u^{4}} \frac{d u}{\cos x} = \int u^{- 4} d u \\ = - \frac{1}{3} u^{- 3} + C = - \frac{1}{3} (\sin x)^{- 3} + C \end{aligned}$

$\int \sin x \sqrt{1 + 3 \cos x} d x$ (6)

$\begin{aligned} \int \sin x \sqrt{1 + 3 \cos x} d x \\ u = 1 + 3 \cos x \Rightarrow \frac{d u}{d x} = - 3 \sin x \Rightarrow d x = \frac{d u}{- 3 \sin x} \\ \int \sin x \sqrt{1 + 3 \cos x} d x = \int \sin x u^{\frac{1}{2}} \frac{d u}{- 3 \sin x} = \int - \frac{1}{3} u^{\frac{1}{2}} d u \\ = - \frac{2}{9} u^{\frac{3}{2}} + C = - \frac{2}{9} \sqrt{(1 + 3 \cos x)^{3}} + C \end{aligned}$

أجد قيمة كل من التكاملات الآتية:

$\int_{1}^{2} \frac{x^{2}}{{(x^{3} + 1)}^{2}} d x$ (7)

$\begin{aligned} \int_{1}^{2} \frac{x^{2}}{{(x^{3} + 1)}^{2}} d x \\ u = x^{3} + 1 \Rightarrow \frac{d u}{d x} = 3 x^{2} \Rightarrow d x = \frac{d u}{3 x^{2}} \\ x = 2 \Rightarrow u = 9 \\ x = 1 \Rightarrow u = 2 \\ \int_{1}^{2} \frac{x^{2}}{{(x^{3} + 1)}^{2}} d x = \int_{2}^{9} \frac{x^{2}}{u^{2}} \frac{d u}{3 x^{2}} = \int_{2}^{9} \frac{1}{3} u^{- 2} d u \\ = - {\frac{1}{3 u} |}_{2}^{9} = - \frac{1}{27} + \frac{1}{6} = \frac{21}{162} \end{aligned}$

$\int_{0}^{1} x \sqrt{3 x^{2} + 2} d x$ (8)

$\begin{aligned} \int_{0}^{1} x \sqrt{3 x^{2} + 2} d x \\ u = 3 x^{2} + 2 \Rightarrow \frac{d u}{d x} = 6 x \Rightarrow d x = \frac{d u}{6 x} \\ x = 1 \Rightarrow u = 5 \\ x = 0 \Rightarrow u = 2 \\ \begin{aligned} \int_{0}^{1} x \sqrt{3 x^{2} + 2} d x = \int_{2}^{5} x u^{\frac{1}{2}} \frac{d u}{6 x} & = \int_{2}^{5} \frac{1}{6} u^{\frac{1}{2}} d u \\ = {\frac{1}{9} u^{\frac{3}{2}} |}_{2}^{5} = \frac{1}{9} \sqrt{125} - \frac{1}{9} \sqrt{8} \end{aligned} \end{aligned}$

$\int_{e}^{e^{2}} \frac{(\ln x)^{2}}{x} d x$ (9)

$\begin{aligned} \int_{e}^{e^{2}} \frac{(\ln x)^{2}}{x} d x \\ u = \ln x \Rightarrow \frac{d u}{d x} = \frac{1}{x} \Rightarrow d x = x d u \\ x = e \Rightarrow u = 1 \\ x = e^{2} \Rightarrow u = 2 \\ \int_{e}^{e^{2}} \frac{(\ln x)^{2}}{x} d x = \int_{1}^{2} \frac{u^{2}}{x} x d u = \int_{1}^{2} u^{2} d u \\ = {\frac{1}{3} u^{3} |}_{1}^{2} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3} \end{aligned}$

$\int_{0}^{1} (x + 1) {(x^{2} + 2 x)}^{5} d x$ (10)

$\begin{aligned} \int_{0}^{1} (x + 1) {(x^{2} + 2 x)}^{5} d x \\ u = x^{2} + 2 x \Rightarrow \frac{d u}{d x} = 2 x + 2 \Rightarrow d x = \frac{d u}{2 x + 2} \\ x = 1 \Rightarrow u = 3 \\ x = 0 \Rightarrow u = 0 \\ \int_{0}^{1} (x + 1) {(x^{2} + 2 x)}^{5} d x = \int_{0}^{3} (x + 1) u^{5} \frac{d u}{2 x + 2} = \int_{0}^{3} \frac{1}{2} u^{5} d u \\ = {\frac{1}{12} u^{6} |}_{0}^{3} = \frac{729}{12} \end{aligned}$

(11) أجد مساحة المنطقة المظللة في التمثيل البياني المجاور.

$\begin{aligned} A = \int_{0}^{2} x \sqrt{x^{2} + 2} d x \\ u = x^{2} + 2 \Rightarrow \frac{d u}{d x} = 2 x \Rightarrow d x = \frac{d u}{2 x} \\ x = 2 \Rightarrow u = 6 \\ x = 0 \Rightarrow u = 2 \\ \begin{matrix} \int_{0}^{2} x \sqrt{x^{2} + 2} d x = \int_{2}^{6} x u^{\frac{1}{2}} \frac{d u}{2 x} = \int_{2}^{6} \frac{1}{2} u^{\frac{1}{2}} d u \\ = {\frac{1}{3} u^{\frac{3}{2}} |}_{2}^{6} = \frac{1}{3} \sqrt{216} - \frac{1}{3} \sqrt{8} \end{matrix} \end{aligned}$

(12) الإيراد الحدي: يمثل الاقتران: $R^{'} (x) = 50 + 3.5 x e^{- 0.1 x^{2}}$ الإيراد الحدي (بالدينار) لكل قطعة تباع من إنتاج إحدى الشركات، حيث $x$ عدد القطع المبيعة، و $R (x)$ إيراد بيع $x$ قطعة بالدينار. أجد اقتران الإيراد $R (x)$ ، علماً بأن $R (0) = 0$ .

$\begin{aligned} R (x) = \int (50 + 3.5 x e^{- 0.1 x^{2}}) d x = \int 50 d x + \int 3.5 x e^{- 0.1 x^{2}} d x \\ = 50 x + \int 3.5 x e^{- 0.1 x^{2}} d x \\ \begin{array}{r} u = - 0.1 x^{2} \Rightarrow \frac{d u}{d x} = - 0.2 x \Rightarrow d x = \frac{d u}{- 0.2 x} \end{array} \\ \begin{aligned} \int (50 + 3.5 x e^{- 0.1 x^{2}}) d x = 50 x + \int 3.5 x e^{u} \frac{d u}{- 0.2 x} \\ = 50 x + \int - 17.5 e^{u} d u \end{aligned} \\ = 50 x - 17.5 e^{- 0.1 x^{2}} + C \\ R (0) = 0 \Rightarrow 0 - 17.5 + C = 0 \Rightarrow C = 17.5 \end{aligned}$

يمثل الاقتران $f^{'} (x)$ في كل مما يأتي ميل المماس لمنحنى الاقتران $f (x)$ المار بالنقطة المعطاة، أستعمل المعلومات المعطاة لإيجاد قاعدة الاقتران $f (x)$ :

$f^{'} (x) = 2 x {(4 x^{2} - 10)}^{2}; (2, 10)$ (13)

$\begin{aligned} f (x) = \int 2 x {(4 x^{2} - 10)}^{2} d x \\ u = 4 x^{2} - 10 \Rightarrow \frac{d u}{d x} = 8 x \Rightarrow d x = \frac{d u}{8 x} \\ \begin{matrix} \int 2 x {(4 x^{2} - 10)}^{2} d x = \int 2 x u^{2} \frac{d u}{8 x} = \int \frac{1}{4} u^{2} d u = \frac{1}{12} u^{3} + C \\ f (2) = 10 \Rightarrow 18 + C = 10 \\ \Rightarrow C = - 8 \\ \Rightarrow f (x) = \frac{1}{12} {(4 x^{2} - 10)}^{3} - 8 \end{matrix} \end{aligned}$

$f^{'} (x) = x^{2} e^{- 0.2 x^{3}}, (0, \frac{3}{2})$ (14)

$\begin{aligned} f (x) = \int x^{2} e^{- 0.2 x^{3}} d x \\ u = - 0.2 x^{3} \Rightarrow \frac{d u}{d x} = - 0.6 x^{2} \Rightarrow d x = \frac{d u}{- 0.6 x^{2}} \\ \int x^{2} e^{- 0.2 x^{3}} d x = \int x^{2} e^{u} \frac{d u}{- 0.6 x^{2}} = \int e^{u} \frac{d u}{- 0.6} \\ \Rightarrow f (x) = - \frac{5}{3} e^{- 0.2 x^{3}} + C \\ f (0) = \frac{3}{2} \Rightarrow - \frac{5}{3} e^{u} d u = - \frac{5}{3} e^{- 0.2 x^{3}} + C \\ \Rightarrow f (x) = - \frac{5}{3} e^{- 0.2 x^{3}} + \frac{19}{6} \end{aligned}$

(15) يتحرك جسيم في مسار مستقيم، وتعطى سرعته المتجهة بالاقتران: $v (t) = \frac{t}{\sqrt{t^{2} + 1}}$ ، حيث $t$ الزمن بالثواني، و $v$ سرعته المتجهة بالمتر لكل ثانية. إذا بدأ الجسيم حركته من نقطة الأصل، فأجد موقعه بعد $t$ ثانية من بدء الحركة.

$\begin{aligned} s (t) = \int \frac{t}{\sqrt{t^{2} + 1}} d t \\ u = t^{2} + 1 \Rightarrow \frac{d u}{d t} = 2 t \Rightarrow d t = \frac{d u}{2 t} \\ \int \frac{t}{\sqrt{t^{2} + 1}} d t = \int u^{- \frac{1}{2}} \frac{d u}{2 t} = \int \frac{1}{2} u^{- \frac{1}{2}} d u = u^{\frac{1}{2}} + C = \sqrt{t^{2} + 1} + C \\ s (t) = \sqrt{t^{2} + 1} + C \\ s (0) = 0 \Rightarrow 1 + C = 0 \Rightarrow C = - 1 \\ s (t) = \sqrt{t^{2} + 1} - 1 \end{aligned}$