មាឌសូលីត និងប្រវែងធ្នូ

មេរៀនទី ៤ ៖ មាឌសូលីត និងប្រវែងធ្នូ

ខ. ដំណោះស្រាយលម្អិតនៃលំហាត់អនុវត្តន៍

ដំណោះស្រាយ

1. \(y = 2x + 1\) លើ \([1, 3]\) ៖

   \(V = \pi \displaystyle\int_1^3 (2x + 1)^2 \, dx\)

   \(= \pi \left[ \frac{4x^3}{3} + 2x^2 + x \right]_1^3 = \frac{158\pi}{3}\)

   ដូចនេះ៖ \(\boxed{V = \frac{158\pi}{3}}\)

2. \(y = x^2 + 1\) លើ \([0, 3]\) ៖

   \(V = \pi \displaystyle\int_0^3 (x^2 + 1)^2 \, dx\)

   \(= \pi \left[ \frac{x^5}{5} + \frac{2x^3}{3} + x \right]_0^3 = \frac{348\pi}{5}\)

   ដូចនេះ៖ \(\boxed{V = \frac{348\pi}{5}}\)

3. \(y = \sqrt{x - 3}\) លើ \([4, 9]\) ៖

   \(V = \pi \displaystyle\int_4^9 (x - 3) \, dx\)

   \(= \pi \left[ \frac{x^2}{2} - 3x \right]_4^9 = \frac{35\pi}{2}\)

   ដូចនេះ៖ \(\boxed{V = \frac{35\pi}{2}}\)

4. \(y = e^x\) លើ \([0, 1]\) ៖

   \(V = \pi \displaystyle\int_0^1 e^{2x} \, dx = \pi \left[ \frac{e^{2x}}{2} \right]_0^1\)

   \(= \frac{\pi(e^2 - 1)}{2}\)

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

5. \(y = \sin x\) លើ \([0, \pi]\) ៖

   \(V = \pi \displaystyle\int_0^\pi \sin^2 x \, dx = \frac{\pi}{2} \left[ x - \frac{\sin 2x}{2} \right]_0^\pi\)

   \(= \frac{\pi^2}{2}\)

   ដូចនេះ៖ \(\boxed{V = \frac{\pi^2}{2}}\)

6. \(y = \dfrac{2}{x}\) លើ \([1, 4]\) ៖

   \(V = 4\pi \displaystyle\int_1^4 \frac{1}{x^2} \, dx = 4\pi \left[ -\frac{1}{x} \right]_1^4\)

   \(= 3\pi\)

   ដូចនេះ៖ \(\boxed{V = 3\pi}\)

ដំណោះស្រាយ

1. \(f(x) = x^2\) និង \(g(x) = 4x - x^2\) ៖

   អាប់ស៊ីសប្រសព្វ ៖ \(4x - x^2 = x^2 \iff x = 0, 2\)

   លើ \([0, 2]\) ៖ \(g(x) \ge f(x) \ge 0\)

   \(V = \pi \displaystyle\int_0^2 ((4x - x^2)^2 - x^4) dx\)

   \(= \pi \left[ \frac{16x^3}{3} - 2x^4 \right]_0^2 = \frac{32\pi}{3}\)

   ដូចនេះ៖ \(\boxed{V = \frac{32\pi}{3}}\)

2. \(f(x) = x^2 - 2x + 3\) និង \(g(x) = 9 - x\) ៖

   អាប់ស៊ីសប្រសព្វ ៖ \(x = -2, 3\)

   លើ \([-2, 3]\) ៖ \(g(x) \ge f(x) \ge 0\)

   \(V = \pi \displaystyle\int_{-2}^3 (-x^4 + 4x^3 - 9x^2 - 6x + 72) dx\)

   \(= \pi \left[ -\frac{x^5}{5} + x^4 - 3x^3 - 3x^2 + 72x \right]_{-2}^3\)

   \(= 250\pi\)

   ដូចនេះ៖ \(\boxed{V = 250\pi}\)

3. \(f(x) = \sqrt{x}\) និង \(g(x) = x^2\) ៖

   អាប់ស៊ីសប្រសព្វ ៖ \(x = 0, 1\)

   លើ \([0, 1]\) ៖ \(\sqrt{x} \ge x^2 \ge 0\)

   \(V = \pi \displaystyle\int_0^1 (x - x^4) dx = \pi \left[ \frac{x^2}{2} - \frac{x^5}{5} \right]_0^1\)

   \(= \frac{3\pi}{10}\)

   ដូចនេះ៖ \(\boxed{V = \frac{3\pi}{10}}\)

4. \(f(x) = 4 - x^2\) និង \(g(x) = x + 2\) ៖

   អាប់ស៊ីសប្រសព្វ ៖ \(x = -2, 1\)

   លើ \([-2, 1]\) ៖ \(4 - x^2 \ge x + 2 \ge 0\)

   \(V = \pi \displaystyle\int_{-2}^1 (x^4 - 9x^2 - 4x + 12) dx\)

   \(= \pi \left[ \frac{x^5}{5} - 3x^3 - 2x^2 + 12x \right]_{-2}^1\)

   \(= \frac{108\pi}{5}\)

   ដូចនេះ៖ \(\boxed{V = \frac{108\pi}{5}}\)

ដំណោះស្រាយ

1. \(x = y^2\) និង \(x = 4\) ជុំវិញ \(Oy\) ៖

   អ័រដោណេប្រសព្វ ៖ \(y = \pm 2\)

   លើ \([-2, 2]\) ៖ \(R = 4, r = y^2\)

   \(V = \pi \displaystyle\int_{-2}^2 (16 - y^4) dy\)

   \(= 2\pi \left[ 16y - \frac{y^5}{5} \right]_0^2 = \frac{256\pi}{5}\)

   ដូចនេះ៖ \(\boxed{V = \frac{256\pi}{5}}\)

2. \(y = \ln x \iff x = e^y\) លើ \([0, 2]\) ៖

   \(V = \pi \displaystyle\int_0^2 (e^y)^2 dy = \pi \displaystyle\int_0^2 e^{2y} dy\)

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

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

ដំណោះស្រាយ

1. គណនា \(F(-1)\) ៖

   \(F(-1) = \displaystyle\int_1^{-1} \sin(\pi t) dt\)

   \(= \left[ -\frac{\cos(\pi t)}{\pi} \right]_1^{-1} = 0\)

   ដូចនេះ៖ \(\boxed{F(-1) = 0}\)

2. គណនា \(F'(0)\) ៖

   តាម FTC ៖ \(F'(x) = \sin(\pi x)\)

   \(\implies F'(0) = \sin(0) = 0\)

   ដូចនេះ៖ \(\boxed{F'(0) = 0}\)

3. គណនា \(F''(x)\) ៖

   \(F''(x) = (F'(x))' = (\sin(\pi x))'\)

   \(= \pi \cos(\pi x)\)

   ដូចនេះ៖ \(\boxed{F''(x) = \pi \cos(\pi x)}\)

4. គណនា \(F''\left(\frac{1}{2}\right)\) ៖

   \(F''\left(\frac{1}{2}\right) = \pi \cos\left( \frac{\pi}{2} \right) = 0\)

   ដូចនេះ៖ \(\boxed{F''\left(\frac{1}{2}\right) = 0}\)

ដំណោះស្រាយ

1. \(y = \frac{x^3}{3} + \frac{1}{4x}\) លើ \([1, 3]\) ៖

   \(y' = x^2 - \frac{1}{4x^2}\)

   \(1 + (y')^2 = \left( x^2 + \frac{1}{4x^2} \right)^2\)

   \(L = \displaystyle\int_1^3 \left( x^2 + \frac{1}{4x^2} \right) dx\)

   \(= \left[ \frac{x^3}{3} - \frac{1}{4x} \right]_1^3 = \frac{53}{6}\)

   ដូចនេះ៖ \(\boxed{L = \frac{53}{6}}\)

2. \(y = \ln(\cos x)\) លើ \([0, \frac{\pi}{3}]\) ៖

   \(y' = -\tan x \implies 1 + (y')^2 = \frac{1}{\cos^2 x}\)

   \(L = \displaystyle\int_0^{\pi/3} \frac{dx}{\cos x} = \Big[ \ln|\tan x + \frac{1}{\cos x}| \Big]_0^{\pi/3}\)

   \(= \ln(2 + \sqrt{3})\)

   ដូចនេះ៖ \(\boxed{L = \ln(2 + \sqrt{3})}\)

3. ប្រៀបធៀប \(AB\) និងធ្នូ \(L\) ចំពោះ \(f(x) = \frac{e^x+e^{-x}}{2}\) ៖

   • \((T): y - f(a) = f'(a)(x - a)\)

   • \(B(0, f(a) - a f'(a))\)

   • \(AB = a\sqrt{1 + [f'(a)]^2} = a f(a)\)

   • \(L = \displaystyle\int_0^a \sqrt{1 + [f'(x)]^2} dx\)

     \(= \frac{e^a - e^{-a}}{2} = f'(a)\)

   • ចំពោះគ្រប់ \(a > 0\) ៖

     \(a\cosh a > \sinh a \implies AB > L\)

     ដូចនេះ៖ \(\boxed{AB > L}\)

ដំណោះស្រាយ

• គណនាមាឌស្វ៊ែរ ៖

  ស្វ៊ែរកាំ \(R\) កើតពីរង្វិលនៃកន្លះរង្វង់ \(y = \sqrt{R^2 - x^2}\) ជុំវិញ \(Ox\) លើ \([-R, R]\)

  តាមវិធីសាស្ត្របន្ទះមូល ៖

  \(V = \pi \displaystyle\int_{-R}^R (R^2 - x^2) dx\)

  \(= 2\pi \left[ R^2 x - \frac{x^3}{3} \right]_0^R = \frac{4}{3}\pi R^3\)

  ដូចនេះ៖ \(\boxed{V = \frac{4}{3}\pi R^3}\)

• គណនាផ្ទៃក្រឡាខាង ៖

  ដេរីវេ ៖ \(y' = \frac{-x}{\sqrt{R^2 - x^2}}\)

  \(\sqrt{1 + (y')^2} = \frac{R}{\sqrt{R^2 - x^2}}\)

  \(y\sqrt{1 + (y')^2} = R\)

  \(S = 2\pi \displaystyle\int_{-R}^R R \, dx = 4\pi R^2\)

  ដូចនេះ៖ \(\boxed{S = 4\pi R^2}\)

រូប (ក) ៖ រង្វិលកន្លះរង្វង់ \(y = \sqrt{R^2 - x^2}\) ជុំវិញ \(Ox\)

រូប (ខ) ៖ ស្វ៊ែរ 3D ដែលមានកាំ \(R\) បង្កើតបានពីអាំងតេក្រាល

ដំណោះស្រាយ

1. កំពូល និងប្រសព្វ \(Ox\) ៖

   • \(y = -(x - 2)^2 + 4 \implies\) \(\boxed{S(2, 4)}\)

   • \(4x - x^2 = 0 \iff x = 0, 4\)

     នាំឱ្យ \(\boxed{O(0, 0) \text{ និង } A(4, 0)}\)

2. ផ្ទៃក្រឡា \(\mathcal{A}\) ៖

   លើ \([0, 4]\) ៖ \(4x - x^2 \ge 0\)

   \(\mathcal{A} = \displaystyle\int_0^4 (4x - x^2) dx = \left[ 2x^2 - \frac{x^3}{3} \right]_0^4\)

   \(= 32 - \frac{64}{3} = \frac{32}{3}\)

   ដូចនេះ៖ \(\boxed{\mathcal{A} = \frac{32}{3}}\)

3. មាឌសូលីតបរិវត្ត \(V\) ៖

   \(V = \pi \displaystyle\int_0^4 (4x - x^2)^2 dx\)

   \(= \pi \displaystyle\int_0^4 (16x^2 - 8x^3 + x^4) dx\)

   \(= \pi \left[ \frac{16x^3}{3} - 2x^4 + \frac{x^5}{5} \right]_0^4 = \frac{512\pi}{15}\)

   ដូចនេះ៖ \(\boxed{V = \frac{512\pi}{15}}\)

រូប (ក) ៖ ផ្ទៃខណ្ឌដោយ \((P): y = 4x - x^2\) និងអ័ក្ស \(Ox\)

រូប (ខ) ៖ សូលីតបរិវត្តដែលកើតពីរង្វិល \((P)\) ជុំវិញអ័ក្ស \(Ox\)

គ. ដំណោះស្រាយលម្អិតនៃលំហាត់ជំពូក

ដំណោះស្រាយ

1. \(\displaystyle I = \int (2x^3 - 3x^2 + 1) dx\)

   \(I = \frac{x^4}{2} - x^3 + x + C\)

   ដូចនេះ៖ \(\boxed{I = \frac{x^4}{2} - x^3 + x + C}\)

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

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

   \(I = \frac{1}{3}\displaystyle\int u^{-1/2} du = \frac{2}{3}\sqrt{x^3 + 1} + C\)

   ដូចនេះ៖ \(\boxed{I = \frac{2}{3}\sqrt{x^3 + 1} + C}\)

3. \(\displaystyle I = \int x e^{x^2} dx\)

   តាង \(u = x^2 \implies du = 2x dx\)

   \(I = \frac{1}{2}\displaystyle\int e^u du = \frac{1}{2}e^{x^2} + C\)

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

4. \(\displaystyle I = \int \sin(3x) \cos(3x) dx\)

   \(I = \frac{1}{2}\displaystyle\int \sin(6x) dx = -\frac{1}{12}\cos(6x) + C\)

   ដូចនេះ៖ \(\boxed{I = -\frac{1}{12}\cos(6x) + C}\)

5. \(\displaystyle I = \int \frac{\ln x}{x} dx\)

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

   \(I = \displaystyle\int u du = \frac{1}{2}\ln^2 x + C\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{2}\ln^2 x + C}\)

6. \(\displaystyle I = \int \frac{1}{x^2 - 4} dx\)

   \(\frac{1}{x^2 - 4} = \frac{1}{4}\left(\frac{1}{x-2} - \frac{1}{x+2}\right)\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{4}\ln\left|\frac{x-2}{x+2}\right| + C}\)

ដំណោះស្រាយ

1. \(\displaystyle I = \int x \cos x \, dx\) ៖

   តាង \(u = x, dv = \cos x dx\)

   \(I = x \sin x - \displaystyle\int \sin x dx\)

   ដូចនេះ៖ \(\boxed{I = x \sin x + \cos x + C}\)

2. \(\displaystyle I = \int x^2 e^x \, dx\) ៖

   តាង \(u = x^2, dv = e^x dx\)

   \(I = x^2 e^x - 2\displaystyle\int x e^x dx\)

   ដូចនេះ៖ \(\boxed{I = (x^2 - 2x + 2)e^x + C}\)

3. \(\displaystyle I = \int \ln x \, dx\) ៖

   តាង \(u = \ln x, dv = dx\)

   \(I = x \ln x - \displaystyle\int dx = x \ln x - x + C\)

   ដូចនេះ៖ \(\boxed{I = x \ln x - x + C}\)

4. \(\displaystyle I = \int x \sqrt{x + 1} \, dx\) ៖

   តាង \(t = \sqrt{x+1} \implies x = t^2-1, dx = 2t dt\)

   \(I = \displaystyle\int (t^2-1) \cdot t \cdot 2t dt = 2\left(\frac{t^5}{5} - \frac{t^3}{3}\right) + C\)

   ដូចនេះ៖ \(\boxed{I = \frac{2}{15}(3x - 2)(x + 1)^{3/2} + C}\)

5. \(\displaystyle I = \int e^x \sin x \, dx\) ៖

   អាំងតេក្រាលវិលជុំ ៖

   \(I = e^x \sin x - e^x \cos x - I\)

   ដូចនេះ៖ \(\boxed{I = \frac{e^x(\sin x - \cos x)}{2} + C}\)

6. \(\displaystyle I = \int \arctan x \, dx\) ៖

   តាង \(u = \arctan x, dv = dx\)

   \(I = x \arctan x - \displaystyle\int \frac{x}{1 + x^2} dx\)

   ដូចនេះ៖ \(\boxed{I = x \arctan x - \frac{1}{2}\ln(1 + x^2) + C}\)

ដំណោះស្រាយ

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

   \(\frac{x^3}{x^2 + 1} = x - \frac{x}{x^2 + 1}\)

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

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

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

   តាមលក្ខណៈ King’s Property ៖

   \(2I = \displaystyle\int_0^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2} \implies I = \frac{\pi}{4}\)

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

3. \(\displaystyle I = \int_0^{\frac{\pi}{4}} \ln(1 + \tan x) \, dx\) ៖

   ជំនួស \(x \to \frac{\pi}{4} - x\) ៖

   \(1 + \tan(\frac{\pi}{4} - x) = \frac{2}{1 + \tan x}\)

   \(2I = \displaystyle\int_0^{\frac{\pi}{4}} \ln 2 \, dx = \frac{\pi}{4}\ln 2\)

   \(\implies I = \frac{\pi}{8}\ln 2\)

   ដូចនេះ៖ \(\boxed{I = \frac{\pi}{8}\ln 2}\)

ដំណោះស្រាយ

1. \(y = x^2 - 4x\) និង \(Ox\) ៖

   \(x^2 - 4x = 0 \iff x = 0, 4\)

   លើ \([0, 4]\) ៖ \(x^2 - 4x \le 0\)

   \(\mathcal{A} = \displaystyle\int_0^4 (4x - x^2) dx = \frac{32}{3}\)

   ដូចនេះ៖ \(\boxed{\mathcal{A} = \frac{32}{3}}\)

2. \(y = x^3 - 3x\) លើ \([-1, 2]\) ៖

   \(x^3 - 3x = 0 \iff x \in \{-\sqrt{3}, 0, \sqrt{3}\}\)

   លើ \([-1, 0]\) ៖ \(y \ge 0\), លើ \([0, \sqrt{3}]\) ៖ \(y \le 0\), លើ \([\sqrt{3}, 2]\) ៖ \(y \ge 0\)

   \(\mathcal{A} = \frac{5}{4} + \frac{9}{4} + \frac{1}{4} = \frac{15}{4}\)

   ដូចនេះ៖ \(\boxed{\mathcal{A} = \frac{15}{4}}\)

3. \(y = -x^2 + 4x\) និង \(y = x\) ៖

   \(-x^2 + 4x = x \iff x = 0, 3\)

   លើ \([0, 3]\) ៖ \(-x^2 + 4x \ge x\)

   \(\mathcal{A} = \displaystyle\int_0^3 (3x - x^2) dx = \frac{9}{2}\)

   ដូចនេះ៖ \(\boxed{\mathcal{A} = \frac{9}{2}}\)

4. \(y = \sqrt{x}\) និង \(y = x^2\) ៖

   \(\sqrt{x} = x^2 \iff x = 0, 1\)

   លើ \([0, 1]\) ៖ \(\sqrt{x} \ge x^2\)

   \(\mathcal{A} = \displaystyle\int_0^1 (\sqrt{x} - x^2) dx = \frac{1}{3}\)

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

ដំណោះស្រាយ

1. \(y = x^2\) និង \(y = 4\) ជុំវិញ \(Ox\) ៖

   អាប់ស៊ីសប្រសព្វ ៖ \(x = \pm 2\)

   \(V = \pi \displaystyle\int_{-2}^2 (16 - x^4) dx\)

   \(= 2\pi \left[ 16x - \frac{x^5}{5} \right]_0^2 = \frac{256\pi}{5}\)

   ដូចនេះ៖ \(\boxed{V = \frac{256\pi}{5}}\)

2. \(y = \sqrt{x}\) និង \(y = x\) ជុំវិញ \(Ox\) ៖

   អាប់ស៊ីសប្រសព្វ ៖ \(x = 0, 1\)

   \(V = \pi \displaystyle\int_0^1 (x - x^2) dx = \frac{\pi}{6}\)

   ដូចនេះ៖ \(\boxed{V = \frac{\pi}{6}}\)

3. \(y = \sin x\) លើ \([0, \pi]\) ជុំវិញ \(Ox\) ៖

   \(V = \pi \displaystyle\int_0^\pi \sin^2 x \, dx\)

   \(= \frac{\pi}{2}\left[ x - \frac{\sin 2x}{2} \right]_0^\pi = \frac{\pi^2}{2}\)

   ដូចនេះ៖ \(\boxed{V = \frac{\pi^2}{2}}\)

ដំណោះស្រាយ

1. \(y = \frac{1}{3}(x^2 + 2)^{3/2}\) ពី \(x = 0\) ដល់ \(3\) ៖

   \(y' = x\sqrt{x^2 + 2} \implies 1 + (y')^2 = (x^2 + 1)^2\)

   \(L = \displaystyle\int_0^3 (x^2 + 1) dx = \left[ \frac{x^3}{3} + x \right]_0^3 = 12\)

   ដូចនេះ៖ \(\boxed{L = 12}\)

2. \(y = \ln(\cos x)\) ពី \(x = 0\) ដល់ \(\frac{\pi}{4}\) ៖

   \(y' = -\tan x \implies 1 + (y')^2 = \frac{1}{\cos^2 x}\)

   \(L = \displaystyle\int_0^{\pi/4} \frac{dx}{\cos x} = \ln(1 + \sqrt{2})\)

   ដូចនេះ៖ \(\boxed{L = \ln(1 + \sqrt{2})}\)

3. \(y = \frac{x^4}{16} + \frac{1}{2x^2}\) ពី \(x = 1\) ដល់ \(2\) ៖

   \(y' = \frac{x^3}{4} - \frac{1}{x^3}\)

   \(1 + (y')^2 = \left( \frac{x^3}{4} + \frac{1}{x^3} \right)^2\)

   \(L = \displaystyle\int_1^2 \left( \frac{x^3}{4} + \frac{1}{x^3} \right) dx = \frac{21}{16}\)

   ដូចនេះ៖ \(\boxed{L = \frac{21}{16}}\)

ដំណោះស្រាយ

1. គណនា \(I_0\) និង \(I_1\) ៖

   • \(I_0 = \displaystyle\int_0^1 1 \, dx = 1 \implies\) \(\boxed{I_0 = 1}\)

   • \(I_1 = \displaystyle\int_0^1 (1 - x^2) dx = \left[ x - \frac{x^3}{3} \right]_0^1 = \frac{2}{3}\)

     នាំឱ្យ \(\boxed{I_1 = \frac{2}{3}}\)

2. បង្ហាញថា \((2n + 1)I_n = 2n I_{n-1}\) ៖

   • តាង \(\begin{cases} u = (1 - x^2)^n \\ dv = dx \end{cases}\)

     \(\implies \begin{cases} du = -2nx(1 - x^2)^{n-1}dx \\ v = x \end{cases}\)

   • \(I_n = \Big[ x(1 - x^2)^n \Big]_0^1 + 2n \displaystyle\int_0^1 x^2(1 - x^2)^{n-1}dx\)

     \(= 2n(I_{n-1} - I_n)\)

   • \(\implies (2n + 1)I_n = 2n I_{n-1}\) (ពិត)

3. ទាញរកតម្លៃ \(I_2\) និង \(I_3\) ៖

   • \(I_2 = \frac{4}{5}I_1 = \frac{4}{5} \cdot \frac{2}{3} = \frac{8}{15} \implies\) \(\boxed{I_2 = \frac{8}{15}}\)

   • \(I_3 = \frac{6}{7}I_2 = \frac{6}{7} \cdot \frac{8}{15} = \frac{16}{35} \implies\) \(\boxed{I_3 = \frac{16}{35}}\)