أتدرب وأحل المسائل

أتدرب وأحل المسائل

التكامل بالأجزاء

أجد كلاً من التكاملات الآتية:

$\int (x + 1) \cos x d x$ (1)

$\begin{array}{ll} u = x + 1 & d v = \cos x d x \\ d u = d x & v = \sin x \\ \int (x + 1) \cos x d x = & (x + 1) \sin x - \int \sin x d x \\ = (x + 1) \sin x + \cos x + C \end{array}$

$\int x e^{x / 2} d x$ (2)

$\begin{aligned} u = x d v = e^{\frac{1}{2} x} d x \\ d u = d x v = 2 e^{\frac{1}{2} x} \\ \int x e^{\frac{1}{2} x} d x = 2 x e^{\frac{1}{2^{x}}} - \int 2 e^{\frac{1}{2} x} d x \\ = 2 x e^{\frac{1}{2} x} - 4 e^{\frac{1}{2} x} + C \end{aligned}$

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

$\begin{array}{lrl} u = 2 x^{2} - 1 & d v & = e^{- x} d x \\ d u = 4 x d x & v & = - e^{- x} \end{array} \int (2 x^{2} - 1) e^{- x} d x = - (2 x^{2} - 1) e^{- x} + \int 4 x e^{- x} d x \begin{array}{ll} u = 4 x & d v = e^{- x} d x \\ d u = 4 d x & v = - e^{- x} \end{array} \begin{aligned} \int (2 x^{2} - 1) e^{- x} d x & = - (2 x^{2} - 1) e^{- x} - 4 x e^{- x} + \int 4 e^{- x} d x \\ = - (2 x^{2} - 1) e^{- x} - 4 x e^{- x} - 4 e^{- x} + C \\ = - e^{- x} (2 x^{2} + 4 x + 3) + C \end{aligned}$

$\int \ln \sqrt{x} d x$ (4)

$\begin{aligned} \int \ln \sqrt{x} d x = \int \frac{1}{2} \ln x d x \\ u = \ln x d v = \frac{1}{2} d x \\ d u = \frac{1}{x} d x v = \frac{1}{2} x \\ \int \frac{1}{2} \ln x d x = \frac{1}{2} x \ln x - \int \frac{1}{2} d x \\ = \frac{1}{2} x \ln x - \frac{1}{2} x + C \end{aligned}$

$\int x \sin x \cos x d x$ (5)

$\begin{aligned} \int x \sin x \cos x d x = \int \frac{1}{2} x \sin 2 x d x \\ u = \frac{1}{2} x d v = \sin 2 x d x \\ d u = \frac{1}{2} d x v = - \frac{1}{2} \cos 2 x \\ \int x \sin x \cos x d x = - \frac{1}{4} x \cos 2 x + \int \frac{1}{4} \cos 2 x d x \\ = - \frac{1}{4} x \cos 2 x + \frac{1}{8} \sin 2 x + C \end{aligned}$

$\int x \sec x \tan x d x$ (6)

$\begin{aligned} u = x d v = \sec x \tan x d x \\ d u = d x v = \sec x \\ \int x \sec x \tan x d x = x \sec x - \int \sec x d x \\ = x \sec x - \int \sec x \times \frac{\sec x + \tan x}{\sec x + \tan x} d x \\ = x \sec x - \int \frac{\sec^{2} x + \sec x \tan x}{\sec x + \tan x} d x \\ = x \sec x - \ln | \sec x + \tan x | + C \end{aligned}$

$\int \frac{x}{\sin^{2} x} d x$ (7)

$\begin{array}{ll} \int \frac{x}{\sin^{2} x} d x = & \int x \csc^{2} x d x \\ u = x & d v = \csc^{2} x d x \\ d u = d x & v = - \cot x \\ \int x \csc^{2} x d x & = - x \cot x + \int \cot x d x \\ = - x \cot x + \int \frac{\cos x}{\sin x} d x \\ = - x \cot x + \ln | \sin x | + C \end{array}$

$\int \frac{\ln x}{x^{3}} d x$ (8)

$\begin{aligned} u = \ln x & d v = x^{- 3} d x \\ d u = \frac{1}{x} d x & v = - \frac{1}{2} x^{- 2} \\ \int x^{- 3} \ln x d x & = - \frac{1}{2} x^{- 2} \ln x - \int - \frac{1}{2} x^{- 2} \frac{1}{x} d x \\ = - \frac{1}{2} x^{- 2} \ln x + \int \frac{1}{2} x^{- 3} d x \\ = - \frac{1}{2} x^{- 2} \ln x - \frac{1}{4} x^{- 2} + C \\ = - \frac{\ln x}{2 x^{2}} - \frac{1}{4 x^{2}} + C \end{aligned}$

$\int 2 x^{2} \sec^{2} x \tan x d x$ (9)

$\begin{array}{ll} u = 2 x^{2} & d v = \sec^{2} x \tan x d x \\ d u = 4 x d x & v = \frac{1}{2} \tan^{2} x \end{array}$

ملاحظة: لإيجاد v استخدمنا طريقة التعويض، حيث: $y = \tan x, d x = \frac{d y}{\sec^{2} x}$ ومنه:

$\begin{aligned} v = \int s e c^{2} x \tan x d x = \int s e c^{2} x y \frac{d y}{s e c^{2} x} = \int y d y = \frac{1}{2} y^{2} = \frac{1}{2} \tan^{2} x \\ \int 2 x^{2} s e c^{2} x \tan x d x = 2 x^{2} (\frac{1}{2} \tan^{2} x) - \int 2 x \tan^{2} x d x \end{aligned} \begin{array}{ll} u = 2 x & d v = \tan^{2} x d x = (s e c^{2} x - 1) d x \\ d u = 2 d x & v = \tan x - x \\ \int 2 x^{2} s e c^{2} x \tan x d x \end{array} \begin{aligned} = x^{2} \tan^{2} x - (2 x (\tan x - x) - \int 2 (\tan x - x) d x) \\ = x^{2} \tan^{2} x - 2 x \tan x + 2 x^{2} + 2 \int (\frac{\sin x}{\cos x} - x) d x \\ = & x^{2} \tan^{2} x - 2 x \tan x + 2 x^{2} - 2 \ln | \cos x | - x^{2} + C \\ = & x^{2} \tan^{2} x - 2 x \tan x + x^{2} - 2 \ln | \cos x | + C \end{aligned}$

$\int (x - 2) \sqrt{8 - x} d x$ (10)

هذه المسألة يمكن حلها بالتعويض، حيث: $(u = \sqrt{8 - x} أو u = 8 - x)$

وحلها بالأجزاء كالآتي:

$\begin{aligned} u = x - 2 d v = (8 - x)^{\frac{1}{2}} d x \\ d u = d x v = - \frac{2}{3} (8 - x)^{\frac{3}{2}} \\ \int (x - 2) \sqrt{8 - x} d x = (x - 2) \times - \frac{2}{3} (8 - x)^{\frac{3}{2}} - \int - \frac{2}{3} (8 - x)^{\frac{3}{2}} d x \\ = - \frac{2}{3} (x - 2) (8 - x)^{\frac{3}{2}} - \frac{4}{15} (8 - x)^{\frac{5}{2}} + C \end{aligned}$

$\int x^{3} \cos 2 x d x$ (11)

بالأجزاء 3 مرات، لنستخدم طريقة الجدول:

حل السؤال 11

$\int x^{3} \cos 2 x d x = \frac{1}{2} x^{3} \sin 2 x + \frac{3}{4} x^{2} \cos 2 x - \frac{3}{4} x \sin 2 x - \frac{3}{8} \cos 2 x + C$

$\int \frac{x}{6^{x}} d x$ (12)

$\begin{aligned} \int \frac{x}{6^{x}} d x = \int x 6^{- x} d x \\ u = x d v = 6^{- x} d x \\ d u = d x v = - \frac{6^{- x}}{\ln 6} \\ \int x 6^{- x} d x = - x \frac{6^{- x}}{\ln 6} + \int \frac{6^{- x}}{\ln 6} d x \\ = - x \frac{6^{- x}}{\ln 6} - \frac{6^{- x}}{(\ln 6)^{2}} + C \end{aligned}$

$\int e^{- x} \sin 2 x d x$ (13)

$\begin{aligned} u = e^{- x} d v = s i n 2 x d x \\ d u = - e^{- x} d x v = \frac{- 1}{2} c o s 2 x \\ \int e^{- x} s i n 2 x d x = - \frac{1}{2} e^{- x} c o s 2 x - \int \frac{1}{2} e^{- x} c o s 2 x d x \end{aligned} \begin{aligned} u = \frac{1}{2} e^{- x} d v = c o s 2 x d x \\ d u = - \frac{1}{2} e^{- x} d x v = \frac{1}{2} s i n 2 x \end{aligned} \begin{aligned} \int e^{- x} s i n 2 x d x = - \frac{1}{2} e^{- x} c o s 2 x - \frac{1}{4} e^{- x} s i n 2 x - \frac{1}{4} \int e^{- x} s i n 2 x d x \\ \int e^{- x} s i n 2 x d x + \frac{1}{4} \int e^{- x} s i n 2 x d x = - \frac{1}{4} e^{- x} (s i n 2 x + 2 c o s 2 x) + C \\ \frac{5}{4} \int e^{- x} s i n 2 x d x = - \frac{1}{4} e^{- x} (s i n 2 x + 2 c o s 2 x) + C \\ \int e^{- x} s i n 2 x d x = - \frac{1}{5} e^{- x} (s i n 2 x + 2 c o s 2 x) + C \end{aligned}$

$\int \cos x \ln \sin x d x$ (14)

$\begin{array}{ll} u = l n s i n x & d v = c o s x d x \\ d u = \frac{c o s x}{s i n x} d x & v = s i n x \end{array} \begin{aligned} \int c o s x l n s i n x d x & = s i n x l n s i n x - \int c o s x d x \\ = s i n x l n s i n x - s i n x + C \end{aligned}$

$\int e^{x} \ln (1 + e^{x}) d x$ (15)

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

أجد قيمة كل من التكاملات الآتية:

$\int_{0}^{π / 2} e^{x} \cos x d x$ (16)

$\begin{aligned} \int e^{x} \cos x d x = \frac{1}{2} e^{x} (\sin x + \cos x) + C \\ \Rightarrow \int_{0}^{\frac{π}{2}} e^{x} \cos x d x = {\frac{1}{2} e^{x} (\sin x + \cos x) |}_{0}^{\frac{π}{2}} \\ = \frac{1}{2} e^{\frac{π}{2}} - \frac{1}{2} e^{0} = \frac{1}{2} e^{\frac{π}{2}} - \frac{1}{2} \end{aligned}$

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

$\begin{aligned} \int_{1}^{e} \ln x^{2} d x = \int_{1}^{e} 2 \ln x d x \\ u = 2 \ln x d v = d x \\ d u = \frac{2}{x} d x v = x \\ \int_{1}^{e} 2 \ln x d x = {2 x \ln x |}_{1}^{e} - \int_{1}^{e} 2 d x \\ = {2 x \ln x |}_{1}^{e} - {2 x |}_{1}^{e} \\ = 2 e \ln e - 2 \ln 1 - 2 e + 2 = 2 e - 0 - 2 e + 2 = 2 \end{aligned}$

$\int_{1}^{2} \ln (x e^{x}) d x$ (18)

$\begin{aligned} \int_{1}^{2} \ln (x e^{x}) d x & = \int_{1}^{2} (\ln x + \ln e^{x}) d x \\ = \int_{1}^{2} (\ln x + x) d x = \int_{1}^{2} \ln x d x + \int_{1}^{2} x d x \end{aligned}$

نجد بطريقة $\int_{1}^{2} \ln x d x$ الأجزاء:

$\begin{array}{ll} u = l n x & d v = d x \\ d u = \frac{1}{x} d x & v = x \end{array} \begin{aligned} \int_{1}^{2} l n x d x = {x l n x |}_{1}^{2} - \int_{1}^{2} d x = {x l n x |}_{1}^{2} - {x |}_{1}^{2} & = 2 l n 2 - l n 1 - 2 + 1 \\ = 2 l n 2 - 1 \end{aligned} \begin{aligned} \int_{1}^{2} x d x = {\frac{1}{2} x^{2} |}_{1}^{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \\ \Rightarrow \int_{1}^{2} l n (x e^{x}) d x = 2 l n 2 - 1 + \frac{3}{2} = 2 l n 2 + \frac{1}{2} \end{aligned}$

$\int_{π / 12}^{π / 9} x \sec^{2} 3 x d x$ (19)

$\begin{array}{ll} u = x & d v = s e c^{2} 3 x d x \\ d u = d x & v = \frac{1}{3} t a n 3 x \end{array} \begin{aligned} \int_{\frac{π}{12}}^{\frac{π}{9}} x s e c^{2} 3 x d x & = {\frac{1}{3} x t a n 3 x |}_{\frac{π}{12}}^{\frac{π}{9}} - \int_{\frac{π}{12}}^{\frac{π}{9}} \frac{1}{3} t a n 3 x d x \\ = {\frac{1}{3} x t a n 3 x |}_{\frac{π}{12}}^{\frac{π}{9}} - \int_{\frac{π}{12}}^{\frac{π}{9}} \frac{1}{3} \frac{s i n 3 x}{c o s 3 x} d x \\ = {\frac{1}{3} x t a n 3 x |}_{\frac{π}{12}}^{\frac{π}{9}} + {\frac{1}{9} l n c o s 3 x |}_{\frac{π}{12}}^{\frac{π}{9}} \\ = \frac{π}{27} t a n \frac{π}{3} - \frac{π}{36} t a n \frac{π}{4} + \frac{1}{9} l n c o s \frac{π}{3} - \frac{1}{9} l n c o s \frac{π}{4} \\ = \frac{π \sqrt{3}}{27} - \frac{π}{36} + \frac{1}{9} l n \frac{1}{2} - \frac{1}{9} l n \frac{1}{\sqrt{2}} \end{aligned}$

$\int_{1}^{e} x^{4} \ln x d x$ (20)

$\begin{aligned} u = \ln x d v = x^{4} d x \\ d u = \frac{d x}{x} v = \frac{1}{5} x^{5} \\ \int_{1}^{e} x^{4} \ln x d x = {\frac{1}{5} x^{5} \ln x |}_{1}^{e} - \int_{1}^{e} \frac{1}{5} x^{4} d x \\ = {\frac{1}{5} x^{5} \ln x |}_{1}^{e} - {\frac{1}{25} x^{5} |}_{1}^{e} \\ = \frac{1}{5} e^{5} - 0 - \frac{1}{25} e^{5} + \frac{1}{25} = \frac{4 e^{5} + 1}{25} \end{aligned}$

$\int_{0}^{π / 2} x^{2} \sin x d x$ (21)

نجد $\int x^{2} \sin x d x$ باستخدام طريقة الجدول:

حل السؤال 21

$\begin{matrix} \int x^{2} \sin x d x = - x^{2} \cos x + 2 x \sin x + 2 \cos x + C \\ \Rightarrow \int_{0}^{\frac{π}{2}} x^{2} \sin x d x = - x^{2} \cos x + 2 x \sin x + {2 \cos x |}_{0}^{\frac{π}{2}} \\ = π - 2 \end{matrix}$

$\int_{0}^{1} x (e^{- 2 x} + e^{- x}) d x$ (22)

$\begin{aligned} u = x d v = (e^{- 2 x} + e^{- x}) d x \\ d u = d x v = - \frac{1}{2} e^{- 2 x} - e^{- x} \\ \int_{0}^{1} x (e^{- 2 x} + e^{- x}) d x = - \frac{1}{2} x e^{- 2 x} - {x e^{- x} |}_{0}^{1} - \int_{0}^{1} (- \frac{1}{2} e^{- 2 x} - e^{- x}) d x \\ = - \frac{1}{2} x e^{- 2 x} - {x e^{- x} |}_{0}^{1} - {(\frac{1}{4} e^{- 2 x} + e^{- x}) |}_{0}^{1} \\ = - \frac{1}{2} e^{- 2} - e^{- 1} - \frac{1}{4} e^{- 2} - e^{- 1} + \frac{1}{4} + 1 \\ = - \frac{3}{4} e^{- 2} - 2 e^{- 1} + \frac{5}{4} \end{aligned}$

$\int_{0}^{1} \frac{x e^{x}}{(1 + x)^{2}} d x$ (23)

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

$\int_{0}^{1} x 3^{x} d x$ (24)

$\begin{aligned} u = x d v = 3^{x} d x \\ d u = d x v = \frac{3^{x}}{\ln 3} \\ \int_{0}^{1} x 3^{x} d x = {x \frac{3^{x}}{\ln 3} |}_{0}^{1} - \int_{0}^{1} \frac{3^{x}}{\ln 3} d x \\ = {x \frac{3^{x}}{\ln 3} |}_{0}^{1} - {\frac{3^{x}}{(\ln 3)^{2}} |}_{0}^{1} \\ = \frac{3}{\ln 3} - \frac{3}{(\ln 3)^{2}} + \frac{1}{(\ln 3)^{2}} = \frac{3 \ln 3 - 2}{(\ln 3)^{2}} \end{aligned}$

أجد كلاً من التكاملات الآتية:

$\int x^{3} e^{x^{2}} d x$ (25)

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

~~$\int \cos (\ln x) d x$~~ (26)

$\begin{aligned} y = l n x \Rightarrow \frac{d y}{d x} = \frac{1}{x} \Rightarrow d x = x d y, x = e^{y} \\ \int c o s (l n x) d x = \int x c o s y d y = \int e^{y} c o s y d y \end{aligned} \begin{aligned} \int e^{y} c o s y d y = \frac{1}{2} e^{y} (s i n y + c o s y) + C \\ \Rightarrow \int c o s (l n x) d x = \frac{1}{2} e^{l n x} (s i n l n x + c o s l n x) + C \\ = \frac{1}{2} x (s i n l n x + c o s l n x) + C \end{aligned}$

$\int x^{3} \sin x^{2} d x$ (27)

$\begin{aligned} y = x^{2} \Rightarrow d x = \frac{d y}{2 x} \\ \int x^{3} s i n x^{2} d x = \int x^{3} s i n y \frac{d y}{2 x} = \int \frac{1}{2} x^{2} s i n y d y = \int \frac{1}{2} y s i n y d y \\ u = \frac{1}{2} y d v = s i n y d y \\ d u = \frac{1}{2} d y v = - c o s y \end{aligned} \begin{aligned} \int \frac{1}{2} y s i n y d y & = - \frac{1}{2} y c o s y + \int \frac{1}{2} c o s y d y \\ = - \frac{1}{2} y c o s y + \frac{1}{2} s i n y + C \\ \int x^{3} s i n x^{2} d x & = - \frac{1}{2} x^{2} c o s x^{2} + \frac{1}{2} s i n x^{2} + C \end{aligned}$

$\int e^{\cos x} \sin 2 x d x$ (28)

$\begin{aligned} y = \cos x \Rightarrow d x = \frac{a y}{- \sin x} \\ \int e^{\cos x} \sin 2 x d x = \int e^{y} (2 \sin x \cos x) \frac{d y}{- \sin x} = \int - 2 y e^{y} d y \\ u = - 2 y d v = e^{y} d y \\ d u = - 2 d y v = e^{y} \\ \int - 2 y e^{y} d y = - 2 y e^{y} + \int 2 e^{y} d y \\ = - 2 y e^{y} + 2 e^{y} + C \\ \Rightarrow \int e^{\cos x} \sin 2 x d x = - 2 \cos x e^{\cos x} + 2 e^{\cos x} + C \end{aligned}$

$\int \sin \sqrt{x} d x$ (29)

$\int \frac{x^{3} e^{x^{2}}}{{(x^{2} + 1)}^{2}} d x$ (30)

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

إذا كان الشكل المجاور يمثل منحنى الاقتران: $f (x) = e^{- x} s i n 2 x$ ، حيث: $x \geq 0$ فأجيب عن الأسئلة الثلاثة الآتية تباعاً:

(31) أجد إحداثيي كل من النقطة $A$ ، والنقطة $B$ .

الإحداثيان x للنقطتين A وB هما أصغر حلين موجبين للمعادلة:

$\begin{aligned} f (x) = e^{- x} \sin 2 x = 0 \\ \Rightarrow \sin 2 x = 0 \Rightarrow 2 x = π, 2 π, \dots \\ \Rightarrow x = \frac{π}{2}, π, \dots \\ \Rightarrow A (\frac{π}{2}, 0), B (π, 0) \end{aligned}$

(32) أجد مساحة المنطقة المظللة.

$A = \int_{0}^{\frac{π}{2}} e^{- x} \sin 2 x d x + (- \int_{\frac{π}{2}}^{π} e^{- x} \sin 2 x d x)$

للبسيط سنجد أولاً: $\int e^{- x} \sin 2 x d x$ (التكامل غير المحدود)

$\begin{array}{ll} u = e^{- x} & d v = s i n 2 x d x \\ d u = - e^{- x} d x & v = - \frac{1}{2} c o s 2 x \end{array} \int e^{- x} s i n 2 x d x = - \frac{1}{2} e^{- x} c o s 2 x - \int \frac{1}{2} e^{- x} c o s 2 x d x \begin{aligned} u = \frac{1}{2} e^{- x} d v = c o s 2 x d x \\ d u = - \frac{1}{2} e^{- x} d x v = \frac{1}{2} s i n 2 x \end{aligned} \begin{aligned} \int e^{- x} s i n 2 x d x = - \frac{1}{2} e^{- x} c o s 2 x - \frac{1}{4} e^{- x} s i n 2 x - \frac{1}{4} \int e^{- x} s i n 2 x d x \\ \int e^{- x} s i n 2 x d x + \frac{1}{4} \int e^{- x} s i n 2 x d x = - \frac{1}{2} e^{- x} c o s 2 x - \frac{1}{4} e^{- x} s i n 2 x \\ \frac{5}{4} \int e^{- x} s i n 2 x d x = - \frac{1}{2} e^{- x} c o s 2 x - \frac{1}{4} e^{- x} s i n 2 x + C \\ \int e^{- x} s i n 2 x d x = - \frac{1}{5} e^{- x} (2 c o s 2 x + s i n 2 x) + C \\ \Rightarrow A = - {\frac{1}{5} e^{- x} (2 c o s 2 x + s i n 2 x) |}_{0}^{\frac{π}{2}} + {\frac{1}{5} e^{- x} (2 c o s 2 x + s i n 2 x) |}_{\frac{π}{2}}^{π} \\ = \frac{2}{5} e^{- \frac{π}{2}} + \frac{2}{5} + \frac{2}{5} e^{- π} + \frac{2}{5} e^{- \frac{π}{2}} \\ = \frac{2}{5} (1 + e^{- π} + 2 e^{- \frac{π}{2}}) \end{aligned}$

(33) يتحرك جسيم في مسار مستقيم، وتعطى سرعته المتجهة بالاقتران: $v (t) = t e^{- t / 2}$ ، حيث $t$ الزمن بالثواني، و $v$ سرعته المتجهة بالمتر لكل ثانية. إذا بدأ الجسيم الحركة من نقطة الأصل، فأجد موقعه بعد $t$ ثانية.

$\begin{aligned} s (t) = \int t e^{- \frac{t}{2}} d t \\ u = t d v = e^{- \frac{t}{2}} d t \\ d u = d t v = - 2 e^{- \frac{t}{2}} \\ s (t) = - 2 t e^{- \frac{t}{2}} - \int - 2 e^{- \frac{t}{2}} d t = - 2 t e^{- \frac{t}{2}} - 4 e^{- \frac{t}{2}} + C \\ s (0) = 0 - 4 + C \\ 0 = 0 - 4 + C \Rightarrow C = 4 \\ \Rightarrow s (t) = - 2 t e^{- \frac{t}{2}} - 4 e^{- \frac{t}{2}} + 4 \end{aligned}$

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

$f^{'} (x) = (x + 2) \sin x; (0, 2)$ (34)

$\begin{aligned} f (x) = \int (x + 2) \sin x d x \\ u = x + 2 d v = \sin x d x \\ d u = d x v = - \cos x \\ f (x) = - (x + 2) \cos x + \int \cos x d x \\ = - (x + 2) \cos x + \sin x + C \\ f (0) = - 2 + 0 + C \\ 2 = - 2 + 0 + C \Rightarrow C = 4 \\ f (x) = - (x + 2) \cos x + \sin x + 4 \end{aligned}$

$f^{'} (x) = 2 x e^{- x}; (0, 3)$ (35)

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

(36) دورة تدريبية: تقدمت دعاء لدورة تدريبية متقدمة في الطباعة. إذا كان عدد الكلمات التي تطبعها دعاء في الدقيقة يزداد بمعدل: $N^{'} (t) = (t + 6) e^{- 0.25 t}$ ، حيث $N (t)$ عدد الكلمات التي تطبعها دعاء في الدقيقة بعد $t$ أسبوعاً من التحاقها بالدورة، فأجد $N (t)$ ، علماً بأن دعاء كانت تطبع 40 كلمة في الدقيقة عند بدء الدورة.

$\begin{aligned} N (t) = \int (t + 6) e^{- 0.25 t} d t \\ u = t + 6 d v = e^{- 0.25 t} d t \\ d u = d t v = - 4 e^{- 0.25 t} \\ N (t) = - 4 (t + 6) e^{- 0.25 t} + \int 4 e^{- 0.25 t} d t \\ = - 4 (t + 6) e^{- 0.25 t} - 16 e^{- 0.25 t} + C \\ N (0) = - 24 - 16 + C \\ 40 = - 24 - 16 + C \Rightarrow C = 80 \\ \Rightarrow N (t) = - 4 (t + 6) e^{- 0.25 t} - 16 e^{- 0.25 t} + 80 \end{aligned}$

إعداد : شبكة منهاجي التعليمية

12 / 02 / 2023

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