(a) To solve \(\int_{1}^{a} x^2 \ln x \, dx = 4\), we integrate by parts. Let \(u = \ln x\) and \(dv = x^2 \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^3}{3}\).

Using integration by parts, \(\int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \frac{x^3}{3} \ln x - \frac{1}{9} x^3\).

Evaluating from 1 to \(a\), we have:

\(\left[ \frac{x^3}{3} \ln x - \frac{1}{9} x^3 \right]_1^a = \frac{a^3}{3} \ln a - \frac{1}{9} a^3 - \left( 0 - \frac{1}{9} \right) = \frac{a^3}{3} \ln a - \frac{1}{9} a^3 + \frac{1}{9}\).

Setting this equal to 4 gives:

\(\frac{a^3}{3} \ln a - \frac{1}{9} a^3 + \frac{1}{9} = 4\).

Multiply through by 9 to clear fractions:

\(3a^3 \ln a - a^3 + 1 = 36\).

Rearrange to find:

\(3a^3 \ln a - a^3 = 35\).

\(a^3 (3 \ln a - 1) = 35\).

Thus, \(a = \left( \frac{35}{3 \ln a - 1} \right)^{\frac{1}{3}}\).

(b) Calculate for \(a = 2.4\) and \(a = 2.8\):

For \(a = 2.4\), \(3 \ln 2.4 - 1 \approx 1.632\), \(\frac{35}{1.632} \approx 21.44\), \(21.44^{\frac{1}{3}} \approx 2.77\).

For \(a = 2.8\), \(3 \ln 2.8 - 1 \approx 2.036\), \(\frac{35}{2.036} \approx 17.19\), \(17.19^{\frac{1}{3}} \approx 2.58\).

Since \(2.4 < 2.77\) and \(2.58 < 2.8\), \(a\) lies between 2.4 and 2.8.

(c) Use the iterative formula \(a_{n+1} = \left( \frac{35}{3 \ln a_n - 1} \right)^{\frac{1}{3}}\).

Starting with \(a_1 = 2.6\):

\(a_2 = \left( \frac{35}{3 \ln 2.6 - 1} \right)^{\frac{1}{3}} \approx 2.635\).

\(a_3 = \left( \frac{35}{3 \ln 2.635 - 1} \right)^{\frac{1}{3}} \approx 2.645\).

\(a_4 = \left( \frac{35}{3 \ln 2.645 - 1} \right)^{\frac{1}{3}} \approx 2.640\).

\(a_5 = \left( \frac{35}{3 \ln 2.640 - 1} \right)^{\frac{1}{3}} \approx 2.640\).

Thus, \(a \approx 2.64\) to 2 decimal places.

(a) To solve \(\int_1^a \frac{\ln x}{\sqrt{x}} \, dx = 6\), we start by integrating:

\(\int \frac{\ln x}{\sqrt{x}} \, dx = 2\sqrt{x} \ln x - 4\sqrt{x} + C\).

Substitute the limits and equate to 6:

\(\left[ 2\sqrt{a} \ln a - 4\sqrt{a} \right] - \left[ 2\cdot1\ln 1 - 4\cdot1 \right] = 6\).

Simplify to get:

\(2\sqrt{a} \ln a - 4\sqrt{a} + 4 = 6\).

Rearrange to find \(a = \exp \left( \frac{1}{\sqrt{a}} + 2 \right)\).

(b) Calculate \(a\) for \(a = 9\) and \(a = 11\):

For \(a = 9\), \(9 < 10.31\).

For \(a = 11\), \(11 > 9.99\).

Thus, \(a\) lies between 9 and 11.

(c) Use the iterative process \(a_{n+1} = \exp \left( \frac{1}{\sqrt{a_n}} + 2 \right)\):

Start with an initial guess, say \(a_1 = 10\).

Calculate successive iterations:

\(a_2 = \exp \left( \frac{1}{\sqrt{10}} + 2 \right) \approx 10.1374\)

\(a_3 = \exp \left( \frac{1}{\sqrt{10.1374}} + 2 \right) \approx 10.1156\)

\(a_4 = \exp \left( \frac{1}{\sqrt{10.1156}} + 2 \right) \approx 10.1190\)

Continue until convergence to 2 decimal places: \(a \approx 10.12\).

(a) Differentiate \(y = \sqrt{x} \cos x\) using the product rule: \(\frac{dy}{dx} = \frac{1}{2\sqrt{x}} \cos x - \sqrt{x} \sin x\).

Set the derivative to zero for the minimum point: \(\frac{1}{2\sqrt{x}} \cos x = \sqrt{x} \sin x\).

Simplify to \(\cos x = 2x \sin x\), leading to \(\tan x = \frac{1}{2x}\).

(b) Use the iterative formula \(a_{n+1} = \pi + \arctan\left(\frac{1}{2a_n}\right)\) starting with \(a_1 = 3\).

Calculate: \(a_2 = 3.3067\), \(a_3 = 3.2917\), \(a_4 = 3.2923\).

The value converges to 3.29 correct to 2 decimal places.

(c) The volume of the solid is given by \(\pi \int (\sqrt{x} \cos x)^2 \, dx\).

Use integration by parts and trigonometric identities to find the integral: \(\frac{x^2}{4} + \frac{x}{4} \sin 2x - \frac{1}{4} \int \sin 2x \, dx\).

Complete the integration to obtain \(\frac{1}{4} x^2 + \frac{1}{4} x \sin 2x + \frac{1}{8} \cos 2x\).

Substitute the limits \(x = 0\) and \(x = \frac{3}{2}\pi\) to find the volume: \(\frac{1}{16} \pi (\pi^2 - 4)\).

(i) To solve \(\int_0^a x \cos \frac{1}{3}x \, dx = 3\), use integration by parts:

Let \(u = x\) and \(dv = \cos \frac{1}{3}x \, dx\). Then \(du = dx\) and \(v = 3 \sin \frac{1}{3}x\).

Integration by parts gives:

\(\int u \, dv = uv - \int v \, du\)

\(= x(3 \sin \frac{1}{3}x) - \int 3 \sin \frac{1}{3}x \, dx\)

\(= 3x \sin \frac{1}{3}x + 9 \cos \frac{1}{3}x\).

Substitute limits and equate to 3:

\(3a \sin \frac{a}{3} + 9 \cos \frac{a}{3} - 9 = 3\)

\(3a \sin \frac{a}{3} + 9 \cos \frac{a}{3} = 12\)

\(a \sin \frac{a}{3} = 4 - 3 \cos \frac{a}{3}\)

\(a = \frac{4 - 3 \cos \frac{a}{3}}{\sin \frac{a}{3}}\)

(ii) Calculate values at \(a = 2.5\) and \(a = 3\):

\(f(a) = a \sin \frac{a}{3} + 3 \cos \frac{a}{3} - 4\)

\(f(2.5) = 0.179\)

\(f(3) = -0.173\)

Since \(f(2.5) > 0\) and \(f(3) < 0\), \(a\) lies between 2.5 and 3.

(iii) Use the iterative formula:

\(a_{n+1} = \frac{4 - 3 \cos \frac{1}{3}a_n}{\sin \frac{1}{3}a_n}\)

Starting with \(a_0 = 2.75\):

\(a_1 = 2.73554\)

\(a_2 = 2.73599\)

\(a_3 = 2.73600\)

\(a_4 = 2.73600\)

Thus, \(a \approx 2.736\) correct to 3 decimal places.

(i) Integrate by parts: Let \(u = x\) and \(dv = e^{-\frac{1}{2}x} \, dx\). Then \(du = dx\) and \(v = -2e^{-\frac{1}{2}x}\).

\(\int x e^{-\frac{1}{2}x} \, dx = -2xe^{-\frac{1}{2}x} + 2 \int e^{-\frac{1}{2}x} \, dx\).

\(= -2xe^{-\frac{1}{2}x} - 4e^{-\frac{1}{2}x} + C\).

Evaluate from 0 to \(a\):

\(\left[-2ae^{-\frac{1}{2}a} - 4e^{-\frac{1}{2}a}\right] - \left[-2(0)e^{-\frac{1}{2}(0)} - 4e^{-\frac{1}{2}(0)}\right] = 2\).

\(-2ae^{-\frac{1}{2}a} - 4e^{-\frac{1}{2}a} + 4 = 2\).

\(-2(a+2)e^{-\frac{1}{2}a} = -2\).

\((a+2)e^{-\frac{1}{2}a} = 1\).

\(a+2 = e^{\frac{1}{2}a}\).

\(a = 2 \ln(a+2)\).

(ii) Calculate \(2 \ln(3+2) = 2 \ln 5 \approx 3.21888\) and \(2 \ln(3.5+2) = 2 \ln 5.5 \approx 3.58352\).

Since \(3.21888 < 3.5\) and \(3.58352 > 3\), \(a\) lies between 3 and 3.5.

(iii) Use iteration: \(a_{n+1} = 2 \ln(a_n + 2)\).

Start with \(a_0 = 3\).

\(a_1 = 2 \ln(3 + 2) = 3.21888\).

\(a_2 = 2 \ln(3.21888 + 2) = 3.3434\).

\(a_3 = 2 \ln(3.3434 + 2) = 3.3584\).

\(a_4 = 2 \ln(3.3584 + 2) = 3.3608\).

\(a_5 = 2 \ln(3.3608 + 2) = 3.3612\).

\(a_6 = 2 \ln(3.3612 + 2) = 3.3613\).

Thus, \(a \approx 3.36\) to 2 decimal places.