إجابات كتاب التمارين

إجابات كتاب التمارين

التكامل المحدود

أجد قيمة كل من التكاملات الآتية:

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

$\begin{aligned} \int_{1}^{5} 10 x^{- 2} d x & = - {10 x^{- 1} |}_{1}^{5} \\ = - {\frac{10}{x} |}_{1}^{5} = (- 2) - (- 10) = 8 \end{aligned}$

$\int_{0}^{2} (2 x^{3} - 4 x + 5) d x$ (2)

$\begin{aligned} \int_{0}^{2} (2 x^{3} - 4 x + 5) d x & = {(\frac{1}{2} x^{4} - 2 x^{2} + 5 x) |}_{0}^{2} \\ = 8 - 8 + 10 = 10 \end{aligned}$

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

$\begin{aligned} \int_{1}^{4} \frac{x^{3} + 2 x^{2}}{\sqrt{x}} d x & = \int_{1}^{4} (\frac{x^{3}}{\sqrt{x}} + \frac{2 x^{2}}{\sqrt{x}}) d x \\ = \int_{1}^{4} (x^{\frac{5}{2}} + 2 x^{\frac{3}{2}}) d x \\ = {(\frac{2}{7} x^{\frac{7}{2}} + \frac{4}{5} x^{\frac{5}{2}}) |}_{1}^{4} \\ = (\frac{256}{7} + \frac{128}{5}) - (\frac{2}{7} + \frac{4}{5}) = \frac{254}{7} + \frac{124}{5} = \frac{2138}{35} \end{aligned}$

$\int_{3}^{6} {(x - \frac{3}{x})}^{2} d x$ (4)

$\begin{aligned} \int_{3}^{6} {(x - \frac{3}{x})}^{2} d x & = \int_{3}^{6} (x^{2} - 6 + \frac{9}{x^{2}}) d x \\ = \int_{3}^{6} (x^{2} - 6 + 9 x^{- 2}) d x \\ = {(\frac{1}{3} x^{3} - 6 x - 9 x^{- 1}) |}_{3}^{6} \\ = {(\frac{1}{3} x^{3} - 6 x - \frac{9}{x}) |}_{3}^{6} \\ = (72 - 36 - \frac{3}{2}) - (9 - 18 - 3) \\ = \frac{93}{2} \end{aligned}$

$\int_{0}^{5} (| x + 3 | - 5) d x$ (5)

$\begin{aligned} | x + 3 | = {\begin{array}{c} - x - 3, x < - 3 \\ x + 3, x \geq - 3 \end{array} \\ \int_{0}^{5} (| x + 3 | - 5) d x = \int_{0}^{5} (x + 3 - 5) d x \\ = \int_{0}^{5} (x - 2) d x \\ = {(\frac{1}{2} x^{2} - 2 x) |}_{0}^{5} = (\frac{25}{2} - 10) - (0 - 0) = \frac{5}{2} \end{aligned}$

$\int_{0}^{6} x (6 - x) d x$ (6)

$\begin{aligned} \int_{0}^{6} x (6 - x) d x & = \int_{0}^{6} (6 x - x^{2}) d x \\ = {(3 x^{2} - \frac{1}{3} x^{3}) |}_{0}^{6} = (108 - \frac{216}{3}) - (0) = \frac{108}{3} = 36 \end{aligned}$

$\int_{1}^{2} (6 x - \frac{12}{x^{4}} + 3) d x$ (7)

$\begin{aligned} \int_{1}^{2} (6 x - \frac{12}{x^{4}} + 3) d x & = \int_{1}^{2} (6 x - 12 x^{- 4} + 3) d x \\ = {(3 x^{2} + 4 x^{- 3} + 3 x) |}_{1}^{2} \\ = {(3 x^{2} + \frac{4}{x^{3}} + 3 x) |}_{1}^{2} \\ = (12 + \frac{1}{2} + 6) - (3 + 4 + 3) = \frac{17}{2} \end{aligned}$

$\int_{0}^{7} | 2 x - 1 | d x$ (8)

$\begin{aligned} | 2 x - 1 | = {\begin{matrix} - 2 x + 1, x < \frac{1}{2} \\ 2 x - 1, x \geq \frac{1}{2} \end{matrix} \\ \int_{0}^{7} | 2 x - 1 | d x = \int_{0}^{\frac{1}{2}} (- 2 x + 1) d x + \int_{\frac{1}{2}}^{7} (2 x - 1) d x \\ = {(- x^{2} + x) |}_{0}^{\frac{1}{2}} + {(x^{2} - x) |}_{\frac{1}{2}}^{7} \\ = (- \frac{1}{4} + \frac{1}{2}) - (0) + (49 - 7) - (\frac{1}{4} - \frac{1}{2}) = \frac{85}{2} \end{aligned}$

$\int_{- 3}^{4} | x | d x$ (9)

$\begin{aligned} | x | = {\begin{aligned} - x & , x < 0 \\ x, & x \geq 0 \end{aligned} \\ \begin{aligned} \int_{- 3}^{4} | x | d x & = \int_{- 3}^{0} - x d x + \int_{0}^{4} x d x \\ = - {\frac{1}{2} x^{2} |}_{- 3}^{0} + {\frac{1}{2} x^{2} |}_{0}^{4} \\ = (0) - (- \frac{9}{2}) + (8) - (0) = \frac{25}{2} \end{aligned} \end{aligned}$

$\int_{1}^{2} \frac{x^{2} + x^{3}}{x} d x$ (10)

$\begin{aligned} \int_{1}^{2} \frac{x^{2} + x^{3}}{x} d x & = \int_{1}^{2} (\frac{x^{2}}{x} + \frac{x^{3}}{x}) d x \\ = \int_{1}^{2} (x + x^{2}) d x \\ = {(\frac{1}{2} x^{2} + \frac{1}{3} x^{3}) |}_{1}^{2} = (2 + \frac{8}{3}) - (\frac{1}{2} + \frac{1}{3}) = \frac{3}{2} + \frac{7}{3} = \frac{23}{6} \end{aligned}$

$\int_{3}^{4} (6 x^{2} - 4 x) d x$ (11)

$\begin{aligned} \int_{3}^{4} (6 x^{2} - 4 x) d x & = {(2 x^{3} + 2 x^{2}) |}_{3}^{4} \\ = (128 + 32) - (54 + 18) = 88 \end{aligned}$

$\int_{10}^{10} \frac{x + 1}{x^{2}} d x$ (12)

$\int_{10}^{10} \frac{x + 1}{x^{2}} d x = 0$

إذا كان $6 \int_{- 3}^{2} f (x) d x = 5, \int_{- 3}^{1} f (x) d x = 4, \int_{- 3}^{2} g (x) d x = - 2$ ، فأجد كلّا مما يأتي:

$\int_{2}^{2} f (x) d x$ (13)

$\int_{2}^{2} f (x) d x = 0$

$\int_{1}^{2} (f (x) - 5) d x$ (14)

$\begin{aligned} \int_{1}^{2} (f (x) - 5) d x & = \int_{1}^{2} f (x) d x - \int_{1}^{2} 5 d x \\ = \int_{1}^{- 3} f (x) d x + \int_{- 3}^{2} f (x) d x + \int_{1}^{2} - 5 d x \\ = - 4 + 5 + {(- 5 x) |}_{1}^{2} \\ = 1 + (- 10) - (- 5) \\ = - 4 \end{aligned}$

$\int_{- 3}^{2} (- 2 f (x) + 5 g (x)) d x$ (15)

$\begin{array}{r} \int_{- 3}^{2} (- 2 f (x) + 5 g (x)) d x = - 2 \int_{- 3}^{2} f (x) d x + 5 \int_{- 3}^{2} g (x) d x \\ = - 2 (5) + 5 (- 2) = - 20 \end{array}$

$\int_{2}^{- 3} (g (x) + 2 x) d x$ (16)

$\begin{aligned} \int_{2}^{- 3} (g (x) + 2 x) d x & = \int_{2}^{- 3} g (x) d x + \int_{2}^{- 3} 2 x d x \\ = - (- 2) + {(x^{2}) |}_{2}^{- 3} = 2 + 9 - 4 = 7 \end{aligned}$

$\int_{2}^{- 3} (f (x) + g (x)) d x$ (17)

$\begin{aligned} \int_{2}^{- 3} (f (x) + g (x)) d x & = \int_{2}^{- 3} f (x) d x + \int_{2}^{- 3} g (x) d x \\ = - 5 + 2 = - 3 \end{aligned}$

$\int_{- 3}^{2} (4 f (x) - 3 g (x)) d x$ (18)

$\begin{aligned} \int_{- 3}^{2} (4 f (x) - 3 g (x)) d x & = 4 \int_{- 3}^{2} f (x) d x - 3 \int_{- 3}^{2} g (x) d x \\ = 4 (5) - 3 (- 2) = 26 \end{aligned}$

(19) إذا كان $f (x) = {\begin{cases} x^{2} & , x < 2 \\ 8 - x & , x \geq 2 \end{cases}$ ، فأجد قيمة $\int_{- 3}^{6} f (x) d x$ .

$\begin{aligned} \int_{- 3}^{6} f (x) d x & = \int_{- 3}^{2} f (x) d x + \int_{2}^{6} f (x) d x \\ = \int_{- 3}^{2} x^{2} d x + \int_{2}^{6} (8 - x) d x \\ = {(\frac{1}{3} x^{3}) |}_{- 3}^{2} + {(8 x - \frac{1}{2} x^{2}) |}_{2}^{6} \\ = (\frac{8}{3}) - (- 9) + (48 - 18) - (16 - 2) = \frac{83}{3} \end{aligned}$

(20) سكان: أشارت دراسة إلى أن عدد السكان في إحدى القرى يتغير شهرياً بمعدل يمكن نمذجته بالاقتران: $P^{'} (t) = 5 + 3 t^{2 / 3}$ ، حيث $t$ عدد الأشهر من الآن ، و $P (t)$ عدد السكان، أجد مقدار الزيادة في عدد سكان القرية في الأشهر الثمانية القادمة.

$\begin{aligned} P (t) & = \int_{0}^{8} (5 + 3 t^{\frac{2}{3}}) d t \\ = {(5 t + \frac{9}{5} t^{\frac{5}{3}}) |}_{0}^{8} \\ = (40 + \frac{288}{5}) - (0) \\ = \frac{488}{5} \end{aligned}$

(21) إذا كان: $\int_{2}^{3} (x^{2} - a) d x = 5$ ، فأجد قيمة الثابت $a$ .

$\begin{aligned} \int_{2}^{3} (x^{2} - a) d x = 5 \\ {(\frac{1}{3} x^{3} - a x) |}_{2}^{3} = 5 \\ (9 - 3 a) - (\frac{8}{3} - 2 a) = 5 \\ \frac{17}{3} - α = 5 \\ a = \frac{2}{3} \end{aligned}$

إعداد : شبكة منهاجي التعليمية

08 / 02 / 2023

حمّل تطبيق منهاجي الجديد

النقاشات

حمّل تطبيق منهاجي الجديد

النقاشات

جميع الحقوق محفوظة © 2025