رياضيات فصل ثاني

التوجيهي علمي

الشرح الملخص أوراق العمل حل اسئلة الدرس النتاجات

التكامل بالتعويض الدرس الثاني :

أجد كلاً من التكاملات الآتية :

أتحقق من فهمي صفحة 32 .

$a \int 4 x^{2} \sqrt{x^{3} - 5} d x S o l u t i o n : l e t u = x^{3} - 5 \Rightarrow d x = \frac{d u}{3 x^{2}} \int 4 x^{2} \sqrt{x^{3} - 5} d x = \int 4 x^{2} u^{\frac{1}{2}} \frac{d u}{3 x^{2}} = \frac{4}{3} \int x^{2} u^{\frac{1}{2}} \frac{d u}{x^{2}} = \frac{4}{3} \int u^{\frac{1}{2}} d u = \frac{4}{3} \times \frac{2}{3} u^{\frac{3}{2}} = \frac{8}{9} {(x^{3} - 5)}^{\frac{3}{2}} + c$

$b \int \frac{1}{2 \sqrt{x}} e^{\sqrt{x}} d x S o l u t i o n : l e t u = \sqrt{x} \Rightarrow d x = \frac{d u}{\frac{1}{2 \sqrt{x}}} = 2 \sqrt{x} d u \int \frac{1}{2 \sqrt{x}} e^{\sqrt{x}} d x = \int \frac{1}{2 \sqrt{x}} e^{u} \times 2 \sqrt{x} d u = \int \frac{1}{2 \sqrt{x}} e^{u} \times 2 \sqrt{x} d u = \int e^{u} d u = e^{u} = e^{\sqrt{x}} + c$

$c \int \frac{{(l n x)}^{3}}{x} d x S o l u t i o n : l e t u = l n x \Rightarrow d x = \frac{d u}{\frac{1}{x}} = x d u \int \frac{{(l n x)}^{3}}{x} d x = \int \frac{u^{3}}{x} x d u = \int \frac{u^{3}}{x} x d u = \int u^{3} d u = \frac{u^{4}}{4} = \frac{1}{4} {(l n x)}^{4} + c$

$d \int \frac{c o s (l n x)}{x} d x S o l u t i o n : l e t u = l n x \Rightarrow d x = \frac{d u}{\frac{1}{x}} = x d u \int \frac{c o s (l n x)}{x} d x = \int \frac{c o s u}{x} x d u = \int \frac{c o s u}{x} x d u = \int c o s u d u = s i n u = s i n (l n x) + c$

$e \int c o s^{4} 5 x s i n 5 x d x S o l u t i o n : l e t u = c o s 5 x \Rightarrow d x = \frac{d u}{- 5 s i n 5 x} \int c o s^{4} 5 x s i n 5 x d x = \int u^{4} s i n 5 x \frac{d u}{- 5 s i n 5 x} = - \frac{1}{5} \int u^{4} s i n 5 x \frac{d u}{s i n 5 x} = - \frac{1}{5} \int u^{4} d u = - \frac{1}{25} u^{5} = - \frac{1}{25} {(c o s 5 x)}^{5} + c$

$f \int x 2^{x^{2}} d x S o l u t i o n : l e t u = x^{2} \Rightarrow d x = \frac{d u}{2 x} \int x 2^{x^{2}} d x = \int x 2^{u} \frac{d u}{2 x} = \frac{1}{2} \int x 2^{u} \frac{d u}{x} = \frac{1}{2} \int 2^{u} d u = \frac{2^{u}}{l n 2} = \frac{2^{x^{2}}}{l n 2} + c$

أتحقق من فهمي صفحة 34 .

أجد كلاً من التكاملات الآتية :

$a \int \frac{x}{\sqrt{1 + 2 x}} d x S o l u t i o n : l e t u = \sqrt{2 x + 1} \Rightarrow d x = \frac{d u}{\frac{2}{2 \sqrt{2 x + 1}}} = \sqrt{2 x + 1} d u = u d u \Rightarrow d x = u d u b u t x = \frac{u^{2} - 1}{2} \int \frac{x}{\sqrt{2 x + 1}} d x = \int \frac{x}{u} u d u = \int \frac{x}{u} u d u = \int x d u = = \int \frac{u^{2} - 1}{2} d u = \frac{1}{2} (\frac{1}{3} u^{3} - u) = \frac{1}{6} {(\sqrt{2 x + 1})}^{3} - \sqrt{2 x + 1} + c$ ملاحظة : يمكن حل هذا السؤال (لاحقاً) بالأجزاء

$b \int x^{7} {(x^{4} - 8)}^{3} d x S o l u t i o n : l e t u = x^{4} - 8 \Rightarrow d x = \frac{d u}{4 x^{3}} b u t x^{4} = u - 8 \int x^{7} {(x^{4} - 8)}^{3} d x = \int x^{7} u^{3} \frac{d u}{4 x^{3}} = \frac{1}{4} \int {\overset{x^{4}}{x^{7}} u}^{3} \frac{d u}{x^{3}} = \int x^{4} u^{3} d u b u t x^{4} = u - 8 = \int (u - 8) u^{3} d u = \int (u^{4} - 8 u^{3}) d u = \frac{1}{5} u^{5} - \frac{8}{4} u^{4} = \frac{1}{5} {(x^{4} - 8)}^{5} - \frac{8}{4} {(x^{4} - 8)}^{4} + c$

$c \int \frac{e^{3 x}}{{(1 - e^{x})}^{2}} d x S o l u t i o n : l e t u = 1 - e^{x} \Rightarrow d x = \frac{d u}{- e^{x}} b u t e^{x} = 1 - u \int \frac{e^{3 x}}{{(1 - e^{x})}^{2}} d x = \int \frac{e^{3 x}}{u^{2}} \frac{d u}{- e^{x}} = - \int \frac{{(e^{x})}^{2}}{u^{2}} d u b u t e^{x} = 1 - u = - \int \frac{{(1 - u)}^{2}}{u^{2}} d u = - \int \frac{1 - 2 u + u^{2}}{u^{2}} d u = \int (u^{- 2} - \frac{2}{u} + 1) d u = - \frac{1}{u} - 2 l n u + u = - \frac{1}{1 - e^{x}} - 2 l n (1 - e^{x}) + 1 - e^{x} + c$

أتحقق من فهمي صفحة 35 .

أجد كلاً من التكاملات الآتية :

$a \int \frac{1}{x + \sqrt[3]{x}} d x S o l u t i o n : l e t u = x^{\frac{1}{3}} \Rightarrow u^{3} = x \Rightarrow 3 u^{2} d u = d x b u t x = u^{3} \int \frac{1}{x + x^{\frac{1}{3}}} d x = \int \frac{1}{u^{3} + u} 3 x^{\frac{2}{3}} d u = 3 \int \frac{u^{2}}{u^{3} + u} d u = \frac{3}{2} \int \frac{2 u}{u^{2} + 1} d u = \frac{3}{2} l n (u^{2} + 1) = \frac{3}{2} l n (x^{\frac{2}{3}} + 1) + c T h e f a s t s o l u t i o n : \int \frac{1}{x + x^{\frac{1}{3}}} d x = \int \frac{x^{\frac{- 1}{3}}}{x^{\frac{- 1}{3}} (x + x^{\frac{1}{3}})} d x = \frac{3}{2} \int \frac{\frac{2}{3} x^{\frac{- 1}{3}}}{x^{\frac{2}{3}} + 1} d x = \frac{3}{2} l n (x^{\frac{2}{3}} + 1) + c$

$b \int x \sqrt[3]{{(1 - x)}^{2}} d x S o l u t i o n : l e t u = {(1 - x)}^{\frac{2}{3}} \Rightarrow d x = \frac{d u}{\frac{2}{3} {(1 - x)}^{\frac{- 1}{3}}} = \frac{3}{2} {(1 - x)}^{\frac{1}{3}} d u b u t 1 - x = u^{\frac{3}{2}} \Rightarrow x = 1 - u^{\frac{3}{2}} \int x \sqrt[3]{{(1 - x)}^{2}} d x = \int x u \times \frac{3}{2} {(1 - x)}^{\frac{1}{3}} d u = \frac{3}{2} \int (1 - u^{\frac{3}{2}}) u \times {(u^{\frac{3}{2}})}^{\frac{1}{3}} d u = \frac{3}{2} \int (1 - u^{\frac{3}{2}}) u^{\frac{3}{2}} d u = \frac{3}{2} \int (u^{\frac{3}{2}} - u^{3}) d u = \frac{3}{2} (\frac{2}{5} u^{\frac{5}{2}} - \frac{1}{4} u^{4}) = \frac{3}{5} u^{\frac{5}{2}} - \frac{3}{8} u^{4} = \frac{3}{5} {(1 - x)}^{\frac{10}{6}} - \frac{3}{8} {(1 - x)}^{\frac{8}{3}} + c$

أتحقق من فهمي صفحة 37 .

يُمثل الاقتران p(x) ، سعر القطعة بالدينار ، تستعمل في أجهزة الحاسوب ، حيث x عدد القطع المبيعة بالمئات .

إذا كان $p' (x) = \frac{- 135 x}{\sqrt{9 + x^{2}}}$ هو معدل تغيّر سعر هذه القطعة ، فأجد p(x) . علماً بأن سعر القطعة الواحدة

هو 30 JD عندما يكون عدد القطع المعيبة منها 400 قطعة .

$S o l u t i o n : p' (x) = \frac{- 135 x}{\sqrt{9 + x^{2}}} \Rightarrow p (x) = \int \frac{- 135 x}{\sqrt{9 + x^{2}}} d x l e t u = 9 + x^{2} \Rightarrow d x = \frac{d u}{2 x} \int \frac{- 135 x}{\sqrt{9 + x^{2}}} d x = \int \frac{- 135 x}{\sqrt{u}} \frac{d u}{2 x} = \frac{- 135}{2} \int \frac{x}{\sqrt{u}} \frac{d u}{x} = \frac{- 135}{2} \int u^{- \frac{1}{2}} d u = - 135 u^{\frac{1}{2}} p (x) = - 135 {(9 + x^{2})}^{\frac{1}{2}} + c p (400) = - 135 {(9 + {(400)}^{2})}^{\frac{1}{2}} + c = 30 \Rightarrow c = 30 + 135 \sqrt{(9 + {(400)}^{2})} \approx 54031.5 \Rightarrow p (x) = - 135 \sqrt{9 + x^{2}} + 54031.5$

أتحقق من فهمي صفحة 39 .

أجد كلاً من التكاملات الآتية :

$a \int s i n^{3} x d x S o l u t i o n : l e t s i n^{3} x = s i n^{2} x \times s i n x = (1 - c o s^{2} x) s i n x l e t u = c o s x \Rightarrow d x = \frac{d u}{- s i n x} \int s i n^{3} x d x = \int (1 - c o s^{2} x) s i n x d x = \int (1 - u^{2}) s i n x \frac{d u}{- s i n x} = \int (1 - u^{2}) s i n x \frac{d u}{- s i n x} = - \int (1 - u^{2}) d u = - u + \frac{1}{3} u^{3} = - c o s x + \frac{1}{3} c o s^{3} x + c$

$b \int c o s^{5} x s i n^{2} x d x S o l u t i o n : l e t c o s^{5} x = c o s^{3} x \times c o s^{2} x = c o s^{3} x \times (1 - s i n^{2} x) l e t u = s i n x \Rightarrow d x = \frac{d u}{c o s x} \int c o s^{5} x s i n^{2} x d x = \int c o s^{3} x (1 - s i n^{2} x) s i n^{2} x d x = \int c o s^{3} x (1 - u^{2}) u^{2} \frac{d u}{c o s x} = \int c o s^{2} x (1 - u^{2}) u^{2} d u = \int (1 - u^{2}) (1 - u^{2}) u^{2} d u = \int (1 - 2 u^{2} + u^{4}) u^{2} d u = \int (u^{2} - 2 u^{4} + u^{6}) d u = \frac{1}{3} u^{3} - \frac{2}{5} u^{5} + \frac{1}{7} u^{7} = \frac{1}{3} s i n^{3} x - \frac{2}{5} s i n^{5} x + \frac{1}{7} s i n^{7} x + c$

أتحقق من فهمي صفحة 41 .

أجد كلاً من التكاملات الآتية :

$a \int t a n^{4} x d x S o l u t i o n : l e t t a n^{4} x = t a n^{2} x \times t a n^{2} x = (s e c^{2} x - 1) t a n^{2} x = t a n^{2} x s e c^{2} x - t a n^{2} x l e t u = t a n x \Rightarrow d x = \frac{d u}{s e c^{2} x} \int t a n^{4} x d x = \int (t a n^{2} x s e c^{2} x - t a n^{2} x) d x = \int (t a n^{2} x s e c^{2} x) d x - \int (t a n^{2} x) d x = \int (t a n^{2} x s e c^{2} x) d x - \int (s e c^{2} x - 1) d x = \int (u^{2} s e c^{2} x) \frac{d u}{s e c^{2} x} - t a n x + x = \int (u^{2} s e c^{2} x) \frac{d u}{s e c^{2} x} - t a n x + x = \int u^{2} d u - t a n x + x = \frac{1}{3} u^{3} - t a n x + x = \frac{1}{3} t a n^{3} x - t a n x + x + c$

$b \int c o t^{5} x d x S o l u t i o n : l e t c o t^{5} x = c o t^{2} x \times c o t^{3} x = (c s c^{2} x - 1) c o t^{3} x = c o t^{3} x c s c^{2} x - c o t^{2} x \times c o t x = c o t^{3} x c s c^{2} x - (c s c^{2} x - 1) \times c o t x = c o t^{3} x c s c^{2} x - c o t x c s c^{2} x - c o t x l e t u = c o t x \Rightarrow d x = - \frac{d u}{c s c^{2} x} \int c o t^{5} x d x = \int (c o t^{3} x c s c^{2} x - c o t x c s c^{2} x) d x - \int c o t x d x = - \int u^{3} x c s c^{2} x \frac{d u}{c s c^{2} x} + \int u c s c^{2} x \frac{d u}{c s c^{2} x} - \int \frac{c o s x}{s i n x} d x = - \int u^{3} x c s c^{2} x \frac{d u}{c s c^{2} x} + \int u c s c^{2} x \frac{d u}{c s c^{2} x} - l n (s i n x) = - \int u^{3} d u + \int u d u - l n (s i n x) = - \frac{1}{4} u^{4} + \frac{1}{2} u^{2} - l n (s i n x) = - \frac{1}{4} c o t^{4} x + \frac{1}{2} c o t^{2} x - l n (s i n x) + c$

$c \int s e c^{4} x t a n^{6} x d x S o l u t i o n : l e t s e c^{4} x t a n^{6} x = s e c^{2} x s e c^{2} x t a n^{6} x = s e c^{2} x (t a n^{2} x + 1) t a n^{6} x = s e c^{2} x t a n^{8} x + s e c^{2} x t a n^{6} x l e t u = t a n x \Rightarrow d x = \frac{d u}{s e c^{2} x} \int s e c^{4} x t a n^{6} x d x = \int (s e c^{2} x u^{8}) \frac{d u}{s e c^{2} x} - \int (s e c^{2} x u^{6}) \frac{d u}{s e c^{2} x} = \int (s e c^{2} x u^{8}) \frac{d u}{s e c^{2} x} - \int (s e c^{2} x u^{6}) \frac{d u}{s e c^{2} x} = \int u^{8} d u - \int u^{6} d u = \frac{1}{9} u^{9} + \frac{1}{7} u^{7} = \frac{1}{9} t a n^{9} x + \frac{1}{7} t a n^{7} x + c$

أتحقق من فهمي صفحة 43.

أجد كلاً من التكاملات الآتية :

$a \int_{0}^{2} x {(x + 1)}^{3} d x S o l u t i o n : l e t u = x + 1 \Rightarrow d x = d u x = u - 1 \int x {(x + 1)}^{3} d x = \int x u^{3} d u = \int (u - 1) u^{3} d u = \int (u^{4} - u^{3}) d u = \frac{1}{5} u^{5} - \frac{1}{4} u^{4} = \frac{1}{5} {(x + 1)}^{5} - \frac{1}{4} {(x + 1)}^{4} 2 0 = \frac{1}{5} ({(3)}^{5} - {(1)}^{5}) - \frac{1}{4} ({(3)}^{4} - {(1)}^{4}) = \frac{242}{5} - 20 = \frac{162}{5}$

$b \int_{0}^{\frac{π}{3}} t a n x s e c x \sqrt{s e c x + 2} d x S o l u t i o n : l e t u = s e c x + 2 \Rightarrow d x = \frac{d u}{s e c x t a n x} \int t a n x s e c x \sqrt{s e c x + 2} d x = \int t a n x s e c x u^{\frac{1}{2}} \frac{d u}{s e c x t a n x} = \int t a n x s e c x u^{\frac{1}{2}} \frac{d u}{s e c x t a n x} = \int u^{\frac{1}{2}} d u = \frac{2}{3} u^{\frac{3}{2}} = \frac{2}{3} {(s e c x + 2)}^{\frac{3}{2}} \frac{π}{3} 0 = \frac{2}{3} {(4)}^{\frac{3}{2}} - \frac{2}{3} {(3)}^{\frac{3}{2}} = \frac{16}{3} - 2 \sqrt{3}$

أتدرب وأحل المسائل .

أجد كلاً من التكاملات الآتية :

$1 \int x^{2} {(2 x^{3} + 5)}^{4} d x S o l u t i o n : l e t u = 2 x^{3} + 5 \Rightarrow d x = \frac{d u}{6 x^{2}} \int x^{2} {(2 x^{3} + 5)}^{4} d x = \int x^{2} u^{4} \frac{d u}{6 x^{2}} = \frac{1}{6} \int x^{2} u^{4} \frac{d u}{x^{2}} = \frac{1}{6} \int u^{4} d u = \frac{1}{30} u^{5} = \frac{1}{30} {(2 x^{3} + 5)}^{5} + c$

$2 \int x^{2} \sqrt{x + 3} d x S o l u t i o n : l e t u = x + 3 \Rightarrow d x = d u b u t x = u - 3 x^{2} = u^{2} - 6 u + 9 \int x^{2} \sqrt{x + 3} d x = \int x^{2} u^{\frac{1}{2}} d x = \int (u^{2} - 6 u + 9) u^{\frac{1}{2}} d x = \int (u^{\frac{5}{2}} - 6 u^{\frac{3}{2}} + 9 u^{\frac{1}{2}}) d u = \frac{2}{7} u^{\frac{5}{2}} - \frac{12}{5} u^{\frac{3}{2}} + 6 u^{\frac{1}{2}} = \frac{2}{7} {(x + 3)}^{\frac{5}{2}} - \frac{12}{5} {(x + 3)}^{\frac{3}{2}} + 6 {(x + 3)}^{\frac{1}{2}} + c$

$3 \int x {(x + 2)}^{3} d x S o l u t i o n : l e t u = x + 2 \Rightarrow d x = d u b u t x = u - 2 \int x {(x + 2)}^{3} d x = \int x u^{3} d u = \int (u - 2) u^{3} d u = \int (u^{4} - 2 u^{3}) d u = \frac{1}{5} u^{5} - \frac{2}{4} u^{4}$