حلول أسئلة الصف السادس الإعدادي

حل اسئلة رياضيات - علوم - عربي وجميع الكتب والمواد الأخرى

تمارين (3-4)

1- جد تكاملات كلاً مما يأتي:

- $\begin{aligned} \int (2 x + 5) (x + 1) d x \end{aligned}$

$\begin{aligned} \int (2 x + 5) (x + 1) d x \\ = \int (2 x^{2} + 2 x + 5 x + 5) d x = \int (2 x^{2} + 7 x + 5) d x = \frac{2 x^{3}}{3} + \frac{7 x^{2}}{2} + 5 x + C \end{aligned}$

- $\begin{aligned} \int_{- 1}^{1} (x^{2} + 3) (x - 2) d x \end{aligned}$

$\begin{aligned} \int_{- 1}^{1} (x^{2} + 3) (x - 2) d x \\ = \int_{- 1}^{1} (x^{3} - 2 x^{2} + 3 x - 6) d x = {[\frac{x^{4}}{4} - \frac{2 x^{3}}{3} + \frac{3 x^{2}}{2} - 6 x]}_{- 1}^{- 1} = (\frac{1}{4} - \frac{2}{3} + \frac{3}{2} - 6) - (\frac{1}{4} + \frac{2}{3} + \frac{3}{2} + 6) \\ = \frac{1}{4} - \frac{2}{3} + \frac{3}{2} - 6 - \frac{1}{4} - \frac{2}{3} - \frac{3}{2} - 6) = - \frac{2}{3} - 6 - \frac{2}{3} - 6 = - \frac{4}{3} - 12 = \frac{- 4 - 36}{3} = \frac{- 40}{3} \end{aligned}$

- $\begin{aligned} \int \sqrt{x} (\sqrt{x} + 5) d x \end{aligned}$

$\begin{aligned} \int \sqrt{x} (\sqrt{x} + 5) d x \\ = \int x^{\frac{1}{2}} (x^{\frac{1}{2}} + 5) d x = \int (x + 5 x^{\frac{1}{2}}) d x = \int \frac{x^{2}}{2} + \frac{5 x^{\frac{3}{2}}}{\frac{3}{2}} + c = \frac{x^{2}}{2} + 5 (\frac{2}{3}) x^{\frac{3}{2}} + C = \frac{x^{2}}{2} + \frac{10}{3} x^{\frac{3}{2}} + C \end{aligned}$

- $\begin{aligned} \int_{0}^{4} \sqrt{x} (x + 1)^{2} d x \end{aligned}$

$\begin{aligned} \int_{0}^{4} \sqrt{x} (x + 1)^{2} d x \\ = & \int_{0}^{4} x^{\frac{1}{2}} (x^{2} + 2 x + 1) d x = \int_{0}^{4} (x^{\frac{5}{2}} + 2 x^{\frac{3}{2}} + x^{\frac{1}{2}}) d x = {[\frac{2}{7} x^{\frac{7}{2}} + 2 \frac{2}{5} x^{\frac{5}{2}} + \frac{2}{3} x^{\frac{3}{2}}]}_{0}^{4} = {[\frac{2}{7} x^{\frac{7}{2}} + \frac{4}{5} x^{\frac{5}{2}} + \frac{2}{3} x^{\frac{3}{2}}]}_{0}^{4} \\ = & {[\frac{2}{7} \sqrt{x^{7}} + \frac{4}{5} \sqrt{x^{5}} + \frac{2}{3} \sqrt{x^{3}}]}_{0}^{4} = (\frac{2}{7} \sqrt{(4)^{7}} + \frac{4}{5} \sqrt{(4)^{5}} + \frac{2}{3} \sqrt{(4)^{3}}) - (0) \\ = & (\frac{2}{7} (2)^{7} + \frac{4}{5} (2)^{5} + \frac{2}{3} (2)^{3}) = (\frac{2}{7} (128) + \frac{4}{5} (32) + \frac{2}{3} (8)) \\ = & \frac{256}{7} + \frac{128}{5} + \frac{16}{3} = \frac{3840 + 2688 + 560}{105} = \frac{7088}{105} \end{aligned}$

- $\begin{aligned} \int \sqrt{x} (\sqrt{x} + 2)^{2} d x \end{aligned}$

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

- $\int_{- 1}^{0} \frac{x^{3} - 27}{x - 3} d x$

$\int_{- 1}^{0} \frac{x^{3} - 27}{x - 3} d x \begin{aligned} = \int_{- 1}^{0} \frac{(x - 3) (x^{2} + 3 x + 9)}{x - 3} d x = {[\frac{x^{3}}{3} + \frac{3 x^{2}}{2} + 9 x]}_{- 1}^{0} = (0) - (- \frac{1}{3} + \frac{3}{2} - 9) = \frac{1}{3} - \frac{3}{2} + 9 \\ = \frac{2 - 9 + 54}{6} = \frac{- 7 + 54}{6} = \frac{47}{6} \end{aligned}$

- $\begin{aligned} \int \frac{x^{4} - 1}{x - 1} d x \end{aligned}$

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

- $\int_{0}^{1} \frac{x d x}{\sqrt{x^{2} + 1}}$

$\begin{aligned} \int_{0}^{1} \frac{x d x}{\sqrt{x^{2} + 1}} \Rightarrow \int_{0}^{1} \frac{x d x}{{(x^{2} + 1)}^{\frac{1}{2}}} \\ = \int_{0}^{1} {(x^{2} + 1)}^{- \frac{1}{2}} x d x, \bar{f} (x) = 2 x \\ = \frac{1}{2} \int_{0}^{1} {(x^{2} + 1)}^{\frac{- 1}{2}} x d x = \frac{1}{2} {[\frac{{(x^{2} + 1)}^{\frac{1}{2}}}{\frac{1}{2}}]}_{0}^{1} = {[\sqrt{x^{2} + 1}]}_{0}^{1} = (\sqrt{1 + 1} - (\sqrt{0 + 1}) = \sqrt{2} - 1 \end{aligned}$

- $\int \frac{x^{2} + 1 d x}{\sqrt[3]{x^{3} + 3 x + 1}}$

$\int \frac{x^{2} + 1 d x}{\sqrt[3]{x^{3} + 3 x + 1}} \begin{aligned} = \int \frac{x^{2} + 1 d x}{{(x^{3} + 3 x + 1)}^{\frac{1}{3}}} = \int {(x^{3} + 3 x + 1)}^{\frac{- 1}{3}} (x^{2} + 1) d x \\ = \frac{1}{3} \int {(x^{3} + 3 x + 1)}^{\frac{- 1}{3}} (3 x^{2} + 3) d x = \frac{1}{3} \frac{{(x^{3} + 3 x + 1)}^{\frac{2}{3}}}{\frac{2}{3}} + C \bar{f} (x) = 3 x^{2} + 3 \\ = \frac{1}{3} \cdot \frac{3}{2} {(x^{3} + 3 x + 1)}^{\frac{2}{3}} + C = \frac{1}{2} {(x^{3} + 3 x + 1)}^{\frac{2}{3}} + C \end{aligned}$

- $\int_{0}^{3} \sqrt[3]{(3 x - 1)^{2}} d x$

$\int_{0}^{3} \sqrt[3]{(3 x - 1)^{2}} d x \begin{aligned} = \int_{0} (3 x - 1)^{\frac{2}{3}} d x \bar{f} (x) = 3 \\ = \frac{1}{3} \int_{0}^{3} (3 x - 1)^{\frac{2}{3}} 3 d x = \frac{1}{3} {[\frac{(3 x - 1)^{\frac{5}{3}}}{\frac{5}{3}}]}_{0}^{3} \\ = \frac{1}{3} \cdot \frac{3}{5} {[(3 x - 1)^{\frac{2}{3}}]}_{0}^{3} = \frac{1}{5} {[\sqrt[3]{(3 x - 1)^{5}}]}_{0}^{3} = \frac{1}{5} [\sqrt[3]{(8)^{5}} - \sqrt[3]{(0 - 1)^{5}} \\ = \frac{1}{5} [(2)^{2} + 1] = \frac{1}{5} [32 + 1] = \frac{33}{5} \end{aligned}$

- $\int \frac{\sqrt[3]{x} + 1}{\sqrt[3]{x^{2}}} d x$

$\int \frac{\sqrt[3]{x} + 1}{\sqrt[3]{x^{2}}} d x = \int \frac{x^{\frac{1}{3}} + 1}{x^{\frac{2}{3}}} d x = \int x^{\frac{- 2}{3}} (x^{\frac{1}{3}} + 1) d x = \int (x^{- \frac{1}{3}} + x^{- \frac{2}{3}}) d x = \frac{3}{2} x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} + C$

- $\int \frac{\sqrt[3]{\sqrt{x} - 1}}{\sqrt{x}} d x$

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

- $\int \frac{x^{4}}{\sqrt[5]{a^{2} x^{5} + b^{2}}} d x$

$\int \frac{x^{4}}{\sqrt[5]{a^{2} x^{5} + b^{2}}} d x \begin{aligned} = \int {(a^{2} x^{5} + b^{2})}^{- \frac{1}{5}} x^{3} d x, \bar{f} (x) = 5 a^{2} x^{4} \\ = \frac{1}{5 a^{2}} \int {(a^{2} x^{5} + b^{2})}^{- \frac{1}{5}} 5 a^{2} x^{4} d x = \frac{1}{5 a^{2}} \frac{{(a^{2} x^{5} + b^{2})}^{\frac{4}{5}}}{\frac{4}{5}} + C = \frac{1}{5 a^{2}} \cdot \frac{5}{4} {(a^{2} x^{5} + b^{2})}^{\frac{4}{5}} + C \\ = \frac{1}{4 a^{2}} {(a^{2} x^{5} + b^{2})}^{\frac{4}{5}} + C \end{aligned}$

- $\int_{0}^{8} \sqrt{x^{2} - 14 x + 49} d x$

ملاحظة: $\int_{a}^{b} \sqrt{x^{2}} d x \int_{a}^{b} | x | d x$

$\begin{aligned} \int_{0}^{8} (x - 7)^{2} d x = \int_{0}^{8} | x - 7 | d x \\ | x - 7 | = {\begin{array}{c} x - 7, x \geq 7 \\ - x + 7, x < 7 \\ 8 \end{array} \\ \int_{0}^{7} (- x + 7) d x + \int_{7}^{7} (x - 7) d x = {[\frac{- x^{2}}{2} + 7 x]}_{0}^{7} + {[\frac{x^{2}}{2} - 7 x]}_{7}^{8} \\ = [(\frac{- 49}{2} + 49) - (0)] + [(\frac{64}{2} - 56) - (\frac{49}{2} - 49)] = [\frac{- 49 + 98}{2}] + [(32 - 56) - (\frac{49 - 98}{2})] \\ = [\frac{49}{2}] + [(- 24) - (\frac{- 49}{2})] = (\frac{49}{2}) + [- 24 + \frac{49}{2}] = (\frac{49}{2}) + (\frac{- 48 + 49}{2}) = \frac{49}{2} + \frac{1}{2} = \frac{50}{2} = 25 \end{aligned}$

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

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

- $\int_{- 1}^{1} \sqrt[5]{3 x^{5} - 2 x^{7}} d x$

$\int_{- 1}^{1} \sqrt[5]{3 x^{5} - 2 x^{7}} d x \begin{aligned} = \int_{- 1}^{1} \sqrt[5]{x^{5} (3 - 2 x^{2})} d x = \int_{- 1}^{1} {(3 - 2 x^{2})}^{\frac{1}{5}} x d x \bar{f} (x) = - 4 x \\ = \frac{- 1}{4} \int_{- 1}^{1} {(3 - 2 x^{2})}^{\frac{1}{5}} - 4 x d x = \frac{- 1}{4} {[\frac{{(3 - 2 x^{2})}^{\frac{6}{5}}}{\frac{6}{5}}]}_{- 1}^{1} \\ = - \frac{1}{4} \cdot \frac{5}{6} {[{(3 - 2 x^{2})}^{\frac{6}{5}}]}_{- 1}^{1} = \frac{- 5}{24} {[{(3 - 2 x^{2})}^{\frac{6}{5}}]}_{- 1}^{1} = \frac{- 5}{24} [(3 - 2)^{\frac{6}{5}} - (3 - 2)^{\frac{6}{5}}] = \frac{- 5}{24} [1 - 1] = 0 \end{aligned}$

- $\int \sqrt[3]{2 x^{5} - 7 x^{3}} d x$

$\int \sqrt[3]{2 x^{5} - 7 x^{3}} d x \begin{aligned} = \int \sqrt[3]{x^{3} (2 x^{2} - 7)} d x \\ = \int {(2 x^{2} - 7)}^{\frac{1}{3}} \times d x, \bar{f} (x) = 4 x \\ = \frac{1}{4} \int {(2 x^{2} - 7)}^{\frac{1}{3}} 4 x d x = \frac{1}{4} \frac{{(2 x^{2} - 7)}^{\frac{4}{3}}}{\frac{4}{3}} + C = \frac{1}{4} \frac{3}{4} {(2 x^{2} - 7)}^{\frac{4}{3}} + C = \frac{3}{16} {(2 x^{2} - 7)}^{\frac{4}{3}} + C \end{aligned}$

- $\int_{1}^{b} (13 - 4 x) d x = 9$ جد قيمة b

$\begin{aligned} {[13 x - 2 x^{2}]}_{1}^{b} = 9 \Rightarrow (13 b - 2 b^{2}) - (13 - 2) = 9 \\ = 13 b - 2 b^{2} - 11 - 9 = 0 \Rightarrow 13 b - 2 b^{2} - 20 = 0 \\ (- 2 b^{2} + 13 b - 20 = 0) (- 1) \Rightarrow 2 b^{2} - 13 b + 20 = 0 \Rightarrow (b - 4) (2 b - 5) = 0 \\ إما b = 4 أو b = \frac{5}{2} \end{aligned}$

حلول أسئلة الصف السادس الإعدادي

حل اسئلة رياضيات - علوم - عربي وجميع الكتب والمواد الأخرى

فايسبوك

واتساب

تيليجرام

طباعة

شرح فيديو

حلول أسئلة الصف السادس الإعدادي

تمارين (3-4)

1- جد تكاملات كلاً مما يأتي:

- $\begin{aligned} \int (2 x + 5) (x + 1) d x \end{aligned}$

- $\begin{aligned} \int_{- 1}^{1} (x^{2} + 3) (x - 2) d x \end{aligned}$

- $\begin{aligned} \int \sqrt{x} (\sqrt{x} + 5) d x \end{aligned}$

- $\begin{aligned} \int_{0}^{4} \sqrt{x} (x + 1)^{2} d x \end{aligned}$

- $\begin{aligned} \int \sqrt{x} (\sqrt{x} + 2)^{2} d x \end{aligned}$

- $\int_{- 1}^{0} \frac{x^{3} - 27}{x - 3} d x$

- $\begin{aligned} \int \frac{x^{4} - 1}{x - 1} d x \end{aligned}$

- $\int_{0}^{1} \frac{x d x}{\sqrt{x^{2} + 1}}$

- $\int \frac{x^{2} + 1 d x}{\sqrt[3]{x^{3} + 3 x + 1}}$

- $\int_{0}^{3} \sqrt[3]{(3 x - 1)^{2}} d x$

- $\int \frac{\sqrt[3]{x} + 1}{\sqrt[3]{x^{2}}} d x$

- $\int \frac{\sqrt[3]{\sqrt{x} - 1}}{\sqrt{x}} d x$

- $\int \frac{x^{4}}{\sqrt[5]{a^{2} x^{5} + b^{2}}} d x$

- $\int_{0}^{8} \sqrt{x^{2} - 14 x + 49} d x$

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

- $\int_{- 1}^{1} \sqrt[5]{3 x^{5} - 2 x^{7}} d x$

- $\int \sqrt[3]{2 x^{5} - 7 x^{3}} d x$

- $\int_{1}^{b} (13 - 4 x) d x = 9$ جد قيمة b

حلول أسئلة الصف السادس الإعدادي

مشاركة

شرح فيديو

فيديو شرح درس تمارين (3-4)

النقاشات

حلول أسئلة الصف السادس الإعدادي

تمارين (3-4)

1- جد تكاملات كلاً مما يأتي:

- ∫(2x+5)(x+1)dx

- ∫−11(x2+3)(x−2)dx

- ∫x(x+5)dx

- ∫04x(x+1)2dx

- ∫x(x+2)2dx

- ∫−10x3−27x−3dx

- ∫x4−1x−1dx

- ∫01xdxx2+1

- ∫x2+1dxx3+3x+13

- ∫03(3x−1)23dx

- ∫x3+1x23dx

- ∫x−13xdx

- ∫x4a2x5+b25dx

- ∫08x2−14x+49dx

- ∫dx(2x−3)2

- ∫−113x5−2x75dx

- ∫2x5−7x33dx

- ∫1b(13−4x)dx=9 جد قيمة b

حلول أسئلة الصف السادس الإعدادي

مشاركة

شرح فيديو

فيديو شرح درس تمارين (3-4)

النقاشات

- $\begin{aligned} \int (2 x + 5) (x + 1) d x \end{aligned}$

- $\begin{aligned} \int_{- 1}^{1} (x^{2} + 3) (x - 2) d x \end{aligned}$

- $\begin{aligned} \int \sqrt{x} (\sqrt{x} + 5) d x \end{aligned}$

- $\begin{aligned} \int_{0}^{4} \sqrt{x} (x + 1)^{2} d x \end{aligned}$

- $\begin{aligned} \int \sqrt{x} (\sqrt{x} + 2)^{2} d x \end{aligned}$

- $\int_{- 1}^{0} \frac{x^{3} - 27}{x - 3} d x$

- $\begin{aligned} \int \frac{x^{4} - 1}{x - 1} d x \end{aligned}$

- $\int_{0}^{1} \frac{x d x}{\sqrt{x^{2} + 1}}$

- $\int \frac{x^{2} + 1 d x}{\sqrt[3]{x^{3} + 3 x + 1}}$

- $\int_{0}^{3} \sqrt[3]{(3 x - 1)^{2}} d x$

- $\int \frac{\sqrt[3]{x} + 1}{\sqrt[3]{x^{2}}} d x$

- $\int \frac{\sqrt[3]{\sqrt{x} - 1}}{\sqrt{x}} d x$

- $\int \frac{x^{4}}{\sqrt[5]{a^{2} x^{5} + b^{2}}} d x$

- $\int_{0}^{8} \sqrt{x^{2} - 14 x + 49} d x$

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

- $\int_{- 1}^{1} \sqrt[5]{3 x^{5} - 2 x^{7}} d x$

- $\int \sqrt[3]{2 x^{5} - 7 x^{3}} d x$

- $\int_{1}^{b} (13 - 4 x) d x = 9$ جد قيمة b