التكامل المحدود

أتدرب وأحل المسائل

أتدرب وأحل المسائل

التكامل المحدود

أجد قيمة كل من التكاملات الآتية:

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

$\begin{aligned} \int_{- 1}^{3} 3 x^{2} d x & = {x^{3} |}_{- 1}^{3} \\ = (3)^{3} - (- 1)^{3} \\ = 28 \end{aligned}$

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

$\begin{aligned} \int_{- 3}^{- 2} 6 d x & = {6 x |}_{- 3}^{- 2} \\ = 6 (- 2) - 6 (- 3) \\ = 6 \end{aligned}$

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

$\begin{aligned} \int_{0}^{2} (3 x^{2} + 4 x + 3) & d x = {(x^{3} + 2 x^{2} + 3 x) |}_{0}^{2} \\ = & ((2)^{3} + 2 (2)^{2} + 3 (2)) - ((0)^{3} + 2 (0)^{2} + 3 (0)) \\ = & 22 \end{aligned}$

(4) $\int_{1}^{8} 8 \sqrt[3]{x} d x$

$\begin{aligned} \int_{1}^{8} 8 \sqrt[3]{x} d x & = \int_{1}^{8} 8 x^{\frac{1}{3}} d x \\ = {6 x^{\frac{4}{3}} |}_{1}^{8} \\ = {6 \sqrt[3]{x^{4}} |}_{1}^{8} \\ = 6 \sqrt[3]{8^{4}} - 6 \sqrt[3]{0^{4}} \\ = 96 \end{aligned}$

(5) $\int_{1}^{9} (\sqrt{x} - \frac{4}{\sqrt{x}}) d x$

$\begin{aligned} \int_{1}^{9} (\sqrt{x} - \frac{4}{\sqrt{x}}) d x & = \int_{1}^{9} (x^{\frac{1}{2}} - 4 x^{- \frac{1}{2}}) d x \\ = {(\frac{2}{3} x^{\frac{3}{2}} - 8 x^{\frac{1}{2}}) |}_{1}^{9} \\ = {(\frac{2}{3} \sqrt{x^{3}} - 8 \sqrt{x}) |}_{1}^{9} \\ = (\frac{2}{3} \sqrt{9^{3}} - 8 \sqrt{9}) - (\frac{2}{3} \sqrt{1^{3}} - 8 \sqrt{1}) \\ = \frac{4}{3} \end{aligned}$

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

$\begin{aligned} \int_{- 2}^{3} (- x^{2} + 4 x - 5) d x = {(- \frac{1}{3} x^{3} + 2 x^{2} - 5 x) |}_{- 2}^{3} \\ = (- \frac{1}{3} (3)^{3} + 2 (3)^{2} - 5 (3)) - (- \frac{1}{3} (- 2)^{3} + 2 (- 2)^{2} - 5 (- 2)) \\ = - \frac{80}{3} \end{aligned}$

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

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

(8) $\int_{- 3}^{3} (9 - x^{2}) d x$

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

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

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

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

$\begin{aligned} \int_{1}^{4} x^{3} (\sqrt{x} + \frac{1}{x}) d x & = \int_{1}^{4} x^{3} (x^{\frac{1}{2}} + x^{- 1}) d x \\ = \int_{1}^{4} (x^{\frac{7}{2}} + x^{2}) d x \\ = {(\frac{2}{9} x^{\frac{9}{2}} + \frac{1}{3} x^{3}) |}_{1}^{4} \\ = {(\frac{2}{9} \sqrt{x^{9}} + \frac{1}{3} x^{3}) |}_{1}^{4} \\ = (\frac{2}{9} \sqrt{4^{9}} + \frac{1}{3} (4)^{3}) - (\frac{2}{9} \sqrt{1^{9}} + \frac{1}{3} (1)^{3}) \\ = \frac{1211}{9} \end{aligned}$

(11) $\int_{1}^{8} (x^{1 / 3} - x^{- 1 / 5}) d x$

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

(12) $\int_{1}^{9} (2 + \sqrt{x})^{2} d x$

$\begin{aligned} \int_{1}^{9} (2 + \sqrt{x})^{2} d x & = \int_{1}^{9} (4 + 4 \sqrt{x} + x) d x \\ = \int_{1}^{9} (4 + 4 x^{\frac{1}{2}} + x) d x \\ = {(4 x + \frac{8}{3} x^{\frac{3}{2}} + \frac{1}{2} x^{2}) |}_{1}^{9} \\ = (4 (9) + \frac{8}{3} (9)^{\frac{3}{2}} + \frac{1}{2} (9)^{2}) - (4 (1) + \frac{8}{3} (1)^{\frac{3}{2}} + \frac{1}{2} (1)^{2}) \\ = \frac{424}{3} \end{aligned}$

(13) $\int_{- 1}^{4} | 3 x - 6 | d x$

أعيد تعريف اقتران القيمة المطلقة:

$| 3 x - 6 | = {\begin{cases} 6 - 3 x, & x < 2 \\ 3 x - 6, & x \geq 2 \end{cases}$

بما أن الاقتران تشعب عند 2 ؛ فإنني أجزىء التكامل عنده:

$\begin{aligned} \int_{- 1}^{4} | 3 x - 6 | d x & = \int_{- 1}^{2} (6 - 3 x) d x + \int_{2}^{4} (3 x - 6) d x \\ = {(6 x - \frac{3}{2} x^{2}) |}_{- 1}^{2} + {(\frac{3}{2} x^{2} - 6 x) |}_{2}^{4} \\ = (6 (2) - \frac{3}{2} (2)^{2}) & - (6 (- 1) - \frac{3}{2} (- 1)^{2}) + (\frac{3}{2} (4)^{2} - 6 (4)) - (\frac{3}{2} (2)^{2} - 6 (2)) \\ = \frac{39}{2} \end{aligned}$

(14) $\int_{0}^{3} | x - 2 | d x$

أعيد تعريف اقتران القيمة المطلقة:

$| x - 2 | = {\begin{cases} 2 - x, & x < 2 \\ x - 2, & x \geq 2 \end{cases}$

بما أن الاقتران تشعب عند 2 ؛ فإنني أجزىء التكامل عنده:

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

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

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

إعداد : شبكة منهاجي التعليمية

01 / 02 / 2023

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