អាំងតេក្រាលកំណត់

មេរៀនទី ២ ៖ អាំងតេក្រាលកំណត់

ខ. ដំណោះស្រាយលម្អិតនៃលំហាត់អនុវត្តន៍

ដំណោះស្រាយ

1. \(\displaystyle I = \int_1^2 f(x) \, dx + \int_2^1 f(x) \, dx\)

   តាមលក្ខណៈត្រឡប់ព្រំដែន៖

   \(\displaystyle\int_2^1 f(x) \, dx = -\int_1^2 f(x) \, dx\)

   នាំឱ្យ \(I = \displaystyle\int_1^2 f(x)dx - \int_1^2 f(x)dx = 0\)

   ដូចនេះ៖ \(\boxed{I = 0}\)

2. \(\displaystyle I = \int_1^1 \frac{f(x)}{3} \, dx\)

   ដោយព្រំដែនក្រោមស្មើព្រំដែនលើ (\(a=b=1\))

   ដូចនេះ៖ \(\boxed{I = 0}\)

ដំណោះស្រាយ

1. \(\displaystyle I = \int_1^5 f(x) dx\)

   តាមលក្ខណៈ Chasles ៖

   \(I = \displaystyle\int_1^2 f(x) dx + \int_2^5 f(x) dx = -4 + 6 = 2\)

   ដូចនេះ៖ \(\boxed{I = 2}\)

2. \(\displaystyle J = \int_5^1 [-4f(x)] dx\)

   \(J = 4\displaystyle\int_1^5 f(x) dx = 4(2) = 8\)

   ដូចនេះ៖ \(\boxed{J = 8}\)

3. \(\displaystyle K = \int_1^5 [4f(x) - 2g(x)] dx\)

   តាមលក្ខណៈលីនេអ៊ែរ ៖

   \(K = 4(2) - 2(8) = 8 - 16 = -8\)

   ដូចនេះ៖ \(\boxed{K = -8}\)

ដំណោះស្រាយ

គណនាអាំងតេក្រាល ៖

ដោយ \(d(2x-1) = 2\,dx\) គេបាន៖

\(\displaystyle\int_1^k \frac{dx}{\sqrt{2x-1}} = \Big[ \sqrt{2x-1} \Big]_1^k = \sqrt{2k-1} - 1\)

ទាញរកតម្លៃ \(k\) ៖

តាមសមត្ថកម្ម \(\sqrt{2k-1} - 1 = 2\)

\(\iff \sqrt{2k-1} = 3 \iff 2k - 1 = 9 \iff k = 5\)

ដូចនេះ៖ \(\boxed{k = 5}\)

ដំណោះស្រាយ

1. \(\displaystyle I = \int_1^1 (3x^2 + 1) \, dx\)

   ដោយព្រំដែនក្រោមស្មើព្រំដែនលើ (\(a=b=1\))

   ដូចនេះ៖ \(\boxed{I = 0}\)

2. \(\displaystyle I = \int_1^2 \left(\frac{1}{x^2} - 3\right) dx\)

   ព្រីមីទីវ៖ \(\displaystyle\int (x^{-2} - 3) dx = \left[ -\frac{1}{x} - 3x \right]_1^2\)

   \(I = \left(-\frac{1}{2} - 6\right) - (-1 - 3) = -\frac{13}{2} + 4 = -\frac{5}{2}\)

   ដូចនេះ៖ \(\boxed{I = -\frac{5}{2}}\)

3. \(\displaystyle I = \int_4^9 \frac{2}{\sqrt{x}} \, dx\)

   \(I = \Big[ 4\sqrt{x} \Big]_4^9 = 4(\sqrt{9} - \sqrt{4}) = 4(3 - 2) = 4\)

   ដូចនេះ៖ \(\boxed{I = 4}\)

4. \(\displaystyle I = \int_2^4 \sqrt{x - 2} \, dx\)

   \(I = \left[ \frac{2}{3}(x - 2)^{\frac{3}{2}} \right]_2^4 = \frac{2}{3}(2)^{\frac{3}{2}} = \frac{4\sqrt{2}}{3}\)

   ដូចនេះ៖ \(\boxed{I = \frac{4\sqrt{2}}{3}}\)

5. \(\displaystyle I = \int_2^4 \sqrt[3]{x - 2} \, dx\)

   \(I = \left[ \frac{3}{4}(x - 2)^{\frac{4}{3}} \right]_2^4 = \frac{3}{4}(2^{\frac{4}{3}}) = \frac{3\sqrt[3]{2}}{2}\)

   ដូចនេះ៖ \(\boxed{I = \frac{3\sqrt[3]{2}}{2}}\)

6. \(\displaystyle I = \int_{-4}^0 \frac{1}{\sqrt{4 - x}} \, dx\)

   \(I = \Big[ -2\sqrt{4 - x} \Big]_{-4}^0 = -2(\sqrt{4} - \sqrt{8}) = 4\sqrt{2} - 4\)

   ដូចនេះ៖ \(\boxed{I = 4\sqrt{2} - 4}\)

ដំណោះស្រាយ (ត)

1. \(\displaystyle I = \int_1^5 \frac{1}{\sqrt{3x + 1}} \, dx\)

   \(I = \left[ \frac{2}{3}\sqrt{3x + 1} \right]_1^5 = \frac{2}{3}(\sqrt{16} - \sqrt{4}) = \frac{4}{3}\)

   ដូចនេះ៖ \(\boxed{I = \frac{4}{3}}\)

2. \(\displaystyle I = \int_1^3 5x(x^2 - 7)^3 \, dx\)

   ដោយ \(d(x^2-7) = 2x\,dx\) គេបាន៖

   \(I = \left[ \frac{5}{8}(x^2 - 7)^4 \right]_1^3 = \frac{5}{8}(16 - 1296) = -800\)

   ដូចនេះ៖ \(\boxed{I = -800}\)

3. \(\displaystyle I = \int_0^2 3x^2\sqrt{x^3 + 1} \, dx\)

   \(I = \left[ \frac{2}{3}(x^3 + 1)^{\frac{3}{2}} \right]_0^2 = \frac{2}{3}(27 - 1) = \frac{52}{3}\)

   ដូចនេះ៖ \(\boxed{I = \frac{52}{3}}\)

4. \(\displaystyle I = \int_1^2 \frac{x + 2}{\sqrt{x^2 + 4x + 1}} \, dx\)

   ដោយ \(d(x^2+4x+1) = 2(x+2)dx\) ៖

   \(I = \Big[ \sqrt{x^2 + 4x + 1} \Big]_1^2 = \sqrt{13} - \sqrt{6}\)

   ដូចនេះ៖ \(\boxed{I = \sqrt{13} - \sqrt{6}}\)

5. \(\displaystyle I = \int_1^4 \left(\frac{3 - \sqrt{x}}{\sqrt{x}}\right) dx\)

   \(I = \Big[ 6\sqrt{x} - x \Big]_1^4 = (12 - 4) - (6 - 1) = 3\)

   ដូចនេះ៖ \(\boxed{I = 3}\)

6. \(\displaystyle I = \int_0^1 (3x^2 - 2x)^2 \, dx\)

   \(I = \left[ \frac{9}{5}x^5 - 3x^4 + \frac{4}{3}x^3 \right]_0^1 = \frac{9}{5} - 3 + \frac{4}{3} = \frac{2}{15}\)

   ដូចនេះ៖ \(\boxed{I = \frac{2}{15}}\)

7. \(\displaystyle I = \int_0^1 \frac{x + 1}{(x^2 + 2x + 2)^3} \, dx\)

   \(I = \left[ -\frac{1}{4(x^2 + 2x + 2)^2} \right]_0^1 = -\frac{1}{100} + \frac{1}{16} = \frac{21}{400}\)

   ដូចនេះ៖ \(\boxed{I = \frac{21}{400}}\)

ដំណោះស្រាយ

1. \(\displaystyle I = \int_1^2 3e^{4x} \, dx\)

   \(I = \left[ \frac{3}{4}e^{4x} \right]_1^2 = \frac{3}{4}(e^8 - e^4)\)

   ដូចនេះ៖ \(\boxed{I = \frac{3}{4}(e^8 - e^4)}\)

2. \(\displaystyle I = \int_0^1 4x^3 e^{2x^4} \, dx\)

   \(I = \left[ \frac{1}{2}e^{2x^4} \right]_0^1 = \frac{1}{2}(e^2 - 1)\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{2}(e^2 - 1)}\)

3. \(\displaystyle I = \int_1^4 \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx\)

   \(I = \Big[ 2e^{\sqrt{x}} \Big]_1^4 = 2(e^2 - e)\)

   ដូចនេះ៖ \(\boxed{I = 2(e^2 - e)}\)

4. \(\displaystyle I = \int_1^3 3e^{2x}(e^{-2x} - 1) \, dx\)

   \(\displaystyle\int_1^3 (3 - 3e^{2x}) dx = \left[ 3x - \frac{3}{2}e^{2x} \right]_1^3\)

   \(I = 6 - \frac{3}{2}(e^6 - e^2)\)

   ដូចនេះ៖ \(\boxed{I = 6 - \frac{3}{2}(e^6 - e^2)}\)

5. \(\displaystyle I = \int_0^1 \frac{dx}{2 + 3x}\)

   \(I = \left[ \frac{1}{3}\ln|2+3x| \right]_0^1 = \frac{1}{3}(\ln 5 - \ln 2)\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{3}\ln\frac{5}{2}}\)

ដំណោះស្រាយ

បម្រែបម្រួលប្រាក់ចំណូលសរុប ៖

\(\Delta R = \displaystyle\int_{30}^{40} (180 + 0.2t) \, dt\)

\(= \Big[ 180t + 0.1t^2 \Big]_{30}^{40}\)

ជំនួសតម្លៃព្រំដែន ៖

\(\Delta R = 7360 - 5490 = 1870 \) ពាន់រៀល

ដូចនេះ៖ \(\boxed{\Delta R = 1\,870\,000 \text{ រៀល}}\)

ដំណោះស្រាយ

1. \(\displaystyle I = \int_0^1 \frac{2x^2}{1 + 2x^3} \, dx\)

   តាង \(u = 1 + 2x^3 \implies 2x^2 dx = \frac{1}{3}du\)

   ព្រំដែន៖ \(x = 0 \implies u = 1\); \(x = 1 \implies u = 3\)

   \(I = \displaystyle\frac{1}{3}\int_1^3 \frac{du}{u} = \left[ \frac{1}{3}\ln u \right]_1^3 = \frac{1}{3}\ln 3\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{3}\ln 3}\)

2. \(\displaystyle I = \int_0^{\ln 2} \frac{e^x}{4e^x - 2} \, dx\)

   តាង \(u = 4e^x - 2 \implies e^x dx = \frac{1}{4}du\)

   ព្រំដែន៖ \(x = 0 \implies u = 2\); \(x = \ln 2 \implies u = 6\)

   \(I = \displaystyle\frac{1}{4}\int_2^6 \frac{du}{u} = \left[ \frac{1}{4}\ln u \right]_2^6 = \frac{1}{4}\ln 3\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{4}\ln 3}\)

3. \(\displaystyle I = \int_0^2 \frac{x}{(1 + x^2)^2} \, dx\)

   តាង \(u = 1 + x^2 \implies x dx = \frac{1}{2}du\)

   ព្រំដែន៖ \(x = 0 \implies u = 1\); \(x = 2 \implies u = 5\)

   \(I = \displaystyle\frac{1}{2}\int_1^5 u^{-2} du = \left[ -\frac{1}{2u} \right]_1^5 = \frac{2}{5}\)

   ដូចនេះ៖ \(\boxed{I = \frac{2}{5}}\)

4. \(\displaystyle I = \int_1^e \frac{(\ln x)^2}{x} \, dx\)

   តាង \(u = \ln x \implies du = \frac{1}{x}dx\)

   ព្រំដែន៖ \(x = 1 \implies u = 0\); \(x = e \implies u = 1\)

   \(I = \displaystyle\int_0^1 u^2 du = \left[ \frac{u^3}{3} \right]_0^1 = \frac{1}{3}\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{3}}\)

5. \(\displaystyle I = \int_0^1 x e^{2x^2} \, dx\)

   តាង \(u = 2x^2 \implies x dx = \frac{1}{4}du\)

   ព្រំដែន៖ \(x = 0 \implies u = 0\); \(x = 1 \implies u = 2\)

   \(I = \displaystyle\frac{1}{4}\int_0^2 e^u du = \frac{1}{4}(e^2 - 1)\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{4}(e^2 - 1)}\)

6. \(\displaystyle I = \int_0^{\frac{\pi}{3}} \tan x \, dx\)

   តាង \(u = \cos x \implies du = -\sin x dx\)

   ព្រំដែន៖ \(x = 0 \implies u = 1\); \(x = \frac{\pi}{3} \implies u = \frac{1}{2}\)

   \(I = \displaystyle\int_{\frac{1}{2}}^1 \frac{du}{u} = \Big[ \ln u \Big]_{\frac{1}{2}}^1 = \ln 2\)

   ដូចនេះ៖ \(\boxed{I = \ln 2}\)

7. \(\displaystyle I = \int_1^e \frac{(1 + \ln x)^2}{2x} \, dx\)

   តាង \(u = 1 + \ln x \implies du = \frac{1}{x}dx\)

   ព្រំដែន៖ \(x = 1 \implies u = 1\); \(x = e \implies u = 2\)

   \(I = \displaystyle\frac{1}{2}\int_1^2 u^2 du = \left[ \frac{u^3}{6} \right]_1^2 = \frac{7}{6}\)

   ដូចនេះ៖ \(\boxed{I = \frac{7}{6}}\)

8. \(\displaystyle I = \int_e^{e^2} \frac{1}{x \ln x} \, dx\)

   តាង \(u = \ln x \implies du = \frac{1}{x}dx\)

   ព្រំដែន៖ \(x = e \implies u = 1\); \(x = e^2 \implies u = 2\)

   \(I = \displaystyle\int_1^2 \frac{du}{u} = \Big[ \ln u \Big]_1^2 = \ln 2\)

   ដូចនេះ៖ \(\boxed{I = \ln 2}\)

ដំណោះស្រាយ

1. \(\displaystyle I = \int_0^1 x e^x \, dx\)

   តាង \(\begin{cases} u = x \\ dv = e^x dx \end{cases} \implies \begin{cases} du = dx \\ v = e^x \end{cases}\)

   \(I = \Big[ x e^x \Big]_0^1 - \displaystyle\int_0^1 e^x dx = e - (e - 1) = 1\)

   ដូចនេះ៖ \(\boxed{I = 1}\)

2. \(\displaystyle I = \int_1^e \ln x \, dx\)

   តាង \(\begin{cases} u = \ln x \\ dv = dx \end{cases} \implies \begin{cases} du = \frac{1}{x}dx \\ v = x \end{cases}\)

   \(I = \Big[ x \ln x \Big]_1^e - \displaystyle\int_1^e dx = e - (e - 1) = 1\)

   ដូចនេះ៖ \(\boxed{I = 1}\)

3. \(\displaystyle I = \int_0^{\frac{\pi}{2}} x \cos x \, dx\)

   តាង \(\begin{cases} u = x \\ dv = \cos x dx \end{cases} \implies \begin{cases} du = dx \\ v = \sin x \end{cases}\)

   \(I = \Big[ x \sin x \Big]_0^{\frac{\pi}{2}} - \displaystyle\int_0^{\frac{\pi}{2}} \sin x dx = \frac{\pi}{2} - 1\)

   ដូចនេះ៖ \(\boxed{I = \frac{\pi}{2} - 1}\)

4. \(\displaystyle I = \int_0^1 (2x + 1) e^{2x} \, dx\)

   តាង \(\begin{cases} u = 2x + 1 \\ dv = e^{2x} dx \end{cases} \implies \begin{cases} du = 2dx \\ v = \frac{1}{2}e^{2x} \end{cases}\)

   \(I = \left[ \frac{2x+1}{2}e^{2x} \right]_0^1 - \displaystyle\int_0^1 e^{2x} dx\)

   \(I = \left(\frac{3}{2}e^2 - \frac{1}{2}\right) - \left( \frac{e^2 - 1}{2} \right) = e^2\)

   ដូចនេះ៖ \(\boxed{I = e^2}\)

5. \(\displaystyle I = \int_1^e x^2 \ln x \, dx\)

   តាង \(\begin{cases} u = \ln x \\ dv = x^2 dx \end{cases} \implies \begin{cases} du = \frac{1}{x}dx \\ v = \frac{x^3}{3} \end{cases}\)

   \(I = \left[ \frac{x^3}{3}\ln x \right]_1^e - \displaystyle\frac{1}{3}\int_1^e x^2 dx\)

   \(I = \frac{e^3}{3} - \left[ \frac{x^3}{9} \right]_1^e = \frac{2e^3 + 1}{9}\)

   ដូចនេះ៖ \(\boxed{I = \frac{2e^3 + 1}{9}}\)

ដំណោះស្រាយ (ត)

1. \(\displaystyle I = \int_0^{\pi} x \sin x \, dx\)

   តាង \(\begin{cases} u = x \\ dv = \sin x dx \end{cases} \implies \begin{cases} du = dx \\ v = -\cos x \end{cases}\)

   \(I = \Big[ -x \cos x \Big]_0^{\pi} + \displaystyle\int_0^{\pi} \cos x dx = \pi\)

   ដូចនេះ៖ \(\boxed{I = \pi}\)

2. \(\displaystyle I = \int_0^1 x^2 e^x \, dx\)

   អាំងតេក្រាលដោយផ្នែកពីរដង ៖

   ព្រីមីទីវនៃ \(x^2 e^x\) គឺ \((x^2 - 2x + 2)e^x\)

   \(I = \Big[ (x^2 - 2x + 2)e^x \Big]_0^1 = e - 2\)

   ដូចនេះ៖ \(\boxed{I = e - 2}\)

3. \(\displaystyle I = \int_0^1 \arctan x \, dx\)

   តាង \(\begin{cases} u = \arctan x \\ dv = dx \end{cases} \implies \begin{cases} du = \frac{1}{1+x^2}dx \\ v = x \end{cases}\)

   \(I = \Big[ x \arctan x \Big]_0^1 - \displaystyle\frac{1}{2}\int_0^1 \frac{2x}{1+x^2} dx\)

   \(I = \frac{\pi}{4} - \frac{1}{2}\ln 2\)

   ដូចនេះ៖ \(\boxed{I = \frac{\pi}{4} - \frac{1}{2}\ln 2}\)

4. \(\displaystyle I = \int_0^{\frac{\pi}{2}} e^x \cos x \, dx\)

   អាំងតេក្រាលវិលជុំ ៖

   ព្រីមីទីវនៃ \(e^x \cos x\) គឺ \(\dfrac{e^x(\sin x + \cos x)}{2}\)

   \(I = \left[ \frac{e^x(\sin x + \cos x)}{2} \right]_0^{\frac{\pi}{2}} = \frac{e^{\frac{\pi}{2}} - 1}{2}\)

   ដូចនេះ៖ \(\boxed{I = \frac{e^{\frac{\pi}{2}} - 1}{2}}\)

ដំណោះស្រាយ

1. ដោះស្រាយសំណួរ (a) ៖

   \(\bullet\) គណនា \(I + J\) ៖

   \[\begin{aligned} I + J &= \int_0^1 \left(\frac{e^x - 1}{e^x + 1} + \frac{2}{e^x + 1}\right) dx \\ &= \int_0^1 \frac{e^x + 1}{e^x + 1} \, dx = \int_0^1 1 \, dx = 1 \end{aligned}\]

   ដូចនេះ៖ \(\boxed{I + J = 1}\)

   \(\bullet\) គណនា \(I - J\) ដោយប្តូរអថេរ ៖

   \[I - J = \int_0^1 \frac{e^x - 3}{e^x + 1} \, dx\]

   តាង \(u = e^x + 1 \implies e^x = u - 1\) និង \(dx = \dfrac{du}{u - 1}\) ៖

   \[\frac{e^x - 3}{e^x + 1} \, dx = \frac{u - 4}{u(u - 1)} \, du = \left( \frac{4}{u} - \frac{3}{u - 1} \right) du\]

   ព្រំដែន៖ \(x = 0 \implies u = 2\) និង \(x = 1 \implies u = e + 1\) ៖

   \[\begin{aligned} I - J &= \int_2^{e+1} \left( \frac{4}{u} - \frac{3}{u - 1} \right) du \\ &= \Big[ 4\ln u - 3\ln(u - 1) \Big]_2^{e+1} \\ &= 4\ln(e+1) - 3\ln e - 4\ln 2 \\ &= 4\ln\left(\frac{e+1}{2}\right) - 3 \end{aligned}\]

   ដូចនេះ៖ \(\boxed{I - J = 4\ln\left(\frac{e+1}{2}\right) - 3}\)

   \(\bullet\) ទាញរកតម្លៃនៃ \(I\) និង \(J\) ៖

   បូកសមីការទាំងពីរ៖

   \[2I = (I+J) + (I-J) = 4\ln\left(\frac{e+1}{2}\right) - 2\]

   ដូចនេះ៖ \(\boxed{I = 2\ln\left(\frac{e+1}{2}\right) - 1}\)

   ដកសមីការទាំងពីរ៖

   \[2J = (I+J) - (I-J) = 4 - 4\ln\left(\frac{e+1}{2}\right)\]

   ដូចនេះ៖ \(\boxed{J = 2 - 2\ln\left(\frac{e+1}{2}\right)}\)

ដំណោះស្រាយ (ត)

1. ដោះស្រាយសំណួរ (b) ៖

   \(\bullet\) គណនា \(I_0\) និង \(I_1\) ៖

   \[\begin{aligned} I_0 &= \int_0^1 x^0 e^{-x} \, dx = \Big[ -e^{-x} \Big]_0^1 \\ &= 1 - \frac{1}{e} = \frac{e - 1}{e} \end{aligned}\]

   ដូចនេះ៖ \(\boxed{I_0 = \frac{e - 1}{e}}\)

   ចំពោះ \(I_1 = \displaystyle\int_0^1 x e^{-x} \, dx\) ៖

   តាង \(\begin{cases} u = x \\ dv = e^{-x} dx \end{cases} \implies \begin{cases} du = dx \\ v = -e^{-x} \end{cases}\)

   \[\begin{aligned} I_1 &= \Big[ -x e^{-x} \Big]_0^1 + \int_0^1 e^{-x} \, dx \\ &= -\frac{1}{e} + \left(1 - \frac{1}{e}\right) = \frac{e - 2}{e} \end{aligned}\]

   ដូចនេះ៖ \(\boxed{I_1 = \frac{e - 2}{e}}\)

   \(\bullet\) ទំនាក់ទំនងរវាង \(I_n\) និង \(I_{n-1}\) ៖

   ចំពោះ \(n \ge 1\) តាមអាំងតេក្រាលដោយផ្នែក៖

   តាង \(\begin{cases} u = x^n \\ dv = e^{-x} dx \end{cases} \implies \begin{cases} du = n x^{n-1} dx \\ v = -e^{-x} \end{cases}\)

   \[\begin{aligned} I_n &= \Big[ -x^n e^{-x} \Big]_0^1 - \int_0^1 (-e^{-x}) \cdot n x^{n-1} \, dx \\ &= -\frac{1}{e} + n \int_0^1 x^{n-1} e^{-x} \, dx \\ &= n I_{n-1} - \frac{1}{e} \end{aligned}\]

   ដូចនេះ៖ \(\boxed{I_n = n I_{n-1} - \frac{1}{e}}\)

   \(\bullet\) ទាញរកតម្លៃនៃ \(I_3\) ៖

   \[\begin{aligned} I_2 &= 2 I_1 - \frac{1}{e} = 2\left(\frac{e - 2}{e}\right) - \frac{1}{e} = \frac{2e - 5}{e} \\ I_3 &= 3 I_2 - \frac{1}{e} = 3\left(\frac{2e - 5}{e}\right) - \frac{1}{e} = \frac{6e - 16}{e} \end{aligned}\]

   ដូចនេះ៖ \(\boxed{I_3 = \frac{6e - 16}{e}}\)