To solve \(\int_0^{\frac{1}{4}\pi} x^2 \cos 2x \, dx\), we use integration by parts. Let \(u = x^2\) and \(dv = \cos 2x \, dx\).

Then \(du = 2x \, dx\) and \(v = \frac{1}{2} \sin 2x\).

Applying integration by parts:

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

\(= \frac{1}{2} x^2 \sin 2x - \int \frac{1}{2} \sin 2x \cdot 2x \, dx\)

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

Now, apply integration by parts again to \(\int x \sin 2x \, dx\).

Let \(u = x\) and \(dv = \sin 2x \, dx\).

Then \(du = dx\) and \(v = -\frac{1}{2} \cos 2x\).

\(\int x \sin 2x \, dx = -\frac{1}{2} x \cos 2x + \int \frac{1}{2} \cos 2x \, dx\)

\(= -\frac{1}{2} x \cos 2x + \frac{1}{4} \sin 2x\).

Substitute back:

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

\(= \frac{1}{2} x^2 \sin 2x + \frac{1}{2} x \cos 2x - \frac{1}{4} \sin 2x\).

Evaluate from \(0\) to \(\frac{1}{4}\pi\):

At \(x = \frac{1}{4}\pi\):

\(\frac{1}{2} \left( \frac{1}{4}\pi \right)^2 \sin \left( \frac{1}{2}\pi \right) + \frac{1}{2} \left( \frac{1}{4}\pi \right) \cos \left( \frac{1}{2}\pi \right) - \frac{1}{4} \sin \left( \frac{1}{2}\pi \right)\)

\(= \frac{1}{2} \cdot \frac{1}{16} \pi^2 \cdot 1 + 0 - \frac{1}{4} \cdot 1\)

\(= \frac{1}{32} \pi^2 - \frac{1}{4}\).

At \(x = 0\):

\(0 + 0 - 0 = 0\).

Thus, the integral evaluates to:

\(\frac{1}{32} \pi^2 - \frac{1}{4}\).

Rewriting \(\frac{1}{4}\) as \(\frac{8}{32}\), we have:

\(\frac{1}{32} \pi^2 - \frac{8}{32} = \frac{1}{32}(\pi^2 - 8)\).

To solve \(\int_{1}^{4} x^{-\frac{3}{2}} \ln x \, dx\), we use integration by parts. Let \(u = \ln x\) and \(dv = x^{-\frac{3}{2}} \, dx\).

Then \(du = \frac{1}{x} \, dx\) and \(v = \int x^{-\frac{3}{2}} \, dx = -2x^{-\frac{1}{2}}\).

Applying integration by parts:

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

\(= \left[ -2x^{-\frac{1}{2}} \ln x \right]_{1}^{4} + 2 \int_{1}^{4} x^{-\frac{1}{2}} \, dx\)

Calculate \(\int x^{-\frac{1}{2}} \, dx\):

\(\int x^{-\frac{1}{2}} \, dx = 2x^{\frac{1}{2}}\)

Substitute back:

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

Evaluate at the bounds:

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

\(= \left[ -\ln 4 + 8 \right] - \left[ 0 + 4 \right]\)

\(= -\ln 4 + 8 - 4\)

\(= 4 - \ln 4\)

Thus, the integral evaluates to \(2 - \ln 4\).

(i) To find \(\int \frac{\ln x}{x^3} \, dx\), use integration by parts. Let \(u = \ln x\) and \(dv = \frac{1}{x^3} \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = -\frac{1}{2x^2}\).

Using integration by parts, \(\int u \, dv = uv - \int v \, du\), we have:

\(\int \frac{\ln x}{x^3} \, dx = \left( \ln x \right) \left( -\frac{1}{2x^2} \right) - \int \left( -\frac{1}{2x^2} \right) \left( \frac{1}{x} \right) \, dx\)

\(= -\frac{\ln x}{2x^2} + \int \frac{1}{2x^3} \, dx\)

\(= -\frac{\ln x}{2x^2} - \frac{1}{4x^2} + C\)

(ii) To evaluate \(\int_1^2 \frac{\ln x}{x^3} \, dx\), substitute the limits into the expression obtained:

\(\left[ -\frac{\ln x}{2x^2} - \frac{1}{4x^2} \right]_1^2\)

\(= \left( -\frac{\ln 2}{8} - \frac{1}{16} \right) - \left( -\frac{0}{2} - \frac{1}{4} \right)\)

\(= -\frac{\ln 2}{8} - \frac{1}{16} + \frac{1}{4}\)

\(= -\frac{\ln 2}{8} + \frac{1}{8}\)

\(= \frac{1}{8} - \frac{\ln 2}{8}\)

\(= \frac{1}{8}(1 - \ln 2)\)

Using \(\ln 2 = \frac{1}{2} \ln 4\), we have:

\(= \frac{1}{8}(1 - \frac{1}{2} \ln 4)\)

\(= \frac{1}{16}(2 - \ln 4)\)

\(= \frac{1}{16}(3 - \ln 4)\)

To solve \(\int_{0}^{\frac{1}{6}\pi} x \cos 3x \, dx\), we use integration by parts. Let \(u = x\) and \(dv = \cos 3x \, dx\). Then \(du = dx\) and \(v = \frac{1}{3} \sin 3x\).

Applying integration by parts:

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

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

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

Integrate \(\int \sin 3x \, dx\):

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

Substitute back:

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

Evaluate from \(0\) to \(\frac{1}{6}\pi\):

\(\left[ \frac{1}{3} x \sin 3x + \frac{1}{9} \cos 3x \right]_{0}^{\frac{1}{6}\pi}\)

At \(x = \frac{1}{6}\pi\):

\(\frac{1}{3} \cdot \frac{1}{6}\pi \cdot \sin \frac{1}{2}\pi + \frac{1}{9} \cdot \cos \frac{1}{2}\pi = \frac{1}{18}\pi \cdot 1 + 0 = \frac{1}{18}\pi\)

At \(x = 0\):

\(\frac{1}{3} \cdot 0 \cdot \sin 0 + \frac{1}{9} \cdot \cos 0 = 0 + \frac{1}{9} \cdot 1 = \frac{1}{9}\)

Subtract the results:

\(\frac{1}{18}\pi - \frac{1}{9} = \frac{1}{18}\pi - \frac{2}{18} = \frac{1}{18}(\pi - 2)\)

To solve \(\int_{0}^{\frac{1}{2}\pi} \theta \sin \frac{1}{2} \theta \, d\theta\), we use integration by parts.

Let \(u = \theta\) and \(dv = \sin \frac{1}{2} \theta \, d\theta\).

Then \(du = d\theta\) and \(v = -2 \cos \frac{1}{2} \theta\).

Using integration by parts, \(\int u \, dv = uv - \int v \, du\), we have:

\(\int \theta \sin \frac{1}{2} \theta \, d\theta = -2\theta \cos \frac{1}{2} \theta + 2 \int \cos \frac{1}{2} \theta \, d\theta\).

The integral \(\int \cos \frac{1}{2} \theta \, d\theta\) is:

\(2 \sin \frac{1}{2} \theta\).

Thus, the indefinite integral is:

\(-2\theta \cos \frac{1}{2} \theta + 4 \sin \frac{1}{2} \theta\).

Now, evaluate from 0 to \(\frac{1}{2}\pi\):

At \(\theta = \frac{1}{2}\pi\):

\(-2\left(\frac{1}{2}\pi\right) \cos \frac{1}{4}\pi + 4 \sin \frac{1}{4}\pi = -\pi \frac{1}{\sqrt{2}} + 4 \frac{1}{\sqrt{2}}\).

At \(\theta = 0\):

\(0\).

Thus, the definite integral is:

\(\left(-\pi \frac{1}{\sqrt{2}} + 4 \frac{1}{\sqrt{2}}\right) - 0 = \frac{4 - \pi}{\sqrt{2}}\).

To solve \(\int_{0}^{\frac{1}{2}\pi} x^2 \sin 2x \, dx\), we use integration by parts.

Let \(u = x^2\) and \(dv = \sin 2x \, dx\). Then \(du = 2x \, dx\) and \(v = -\frac{1}{2} \cos 2x\).

Applying integration by parts:

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

\(= -\frac{1}{2} x^2 \cos 2x + \int \frac{1}{2} \cos 2x \cdot 2x \, dx\)

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

Now, integrate \(\int x \cos 2x \, dx\) by parts again:

Let \(u = x\) and \(dv = \cos 2x \, dx\). Then \(du = dx\) and \(v = \frac{1}{2} \sin 2x\).

\(\int x \cos 2x \, dx = \frac{1}{2} x \sin 2x - \int \frac{1}{2} \sin 2x \, dx\)

\(= \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x\)

Substitute back:

\(-\frac{1}{2} x^2 \cos 2x + \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x\)

Evaluate from 0 to \(\frac{1}{2}\pi\):

At \(x = \frac{1}{2}\pi\):

\(-\frac{1}{2} \left( \frac{1}{2}\pi \right)^2 \cdot (-1) + \frac{1}{2} \cdot \frac{1}{2}\pi \cdot 0 + \frac{1}{4} \cdot (-1)\)

\(= \frac{1}{8} \pi^2 - \frac{1}{4}\)

At \(x = 0\):

\(0 + 0 + \frac{1}{4}\)

Subtract the two results:

\(\left( \frac{1}{8} \pi^2 - \frac{1}{4} \right) - \frac{1}{4} = \frac{1}{8} (\pi^2 - 4)\)

To solve \(\int_{0}^{\frac{1}{2}} xe^{-2x} \, dx\), we use integration by parts. Let \(u = x\) and \(dv = e^{-2x} \, dx\). Then \(du = dx\) and \(v = -\frac{1}{2} e^{-2x}\).

Applying integration by parts:

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

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

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

Integrate \(\int e^{-2x} \, dx\):

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

Substitute back:

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

Evaluate from 0 to \(\frac{1}{2}\):

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

\(= \left(-\frac{1}{4} e^{-1} - \frac{1}{4} e^{-1}\right) - \left(0 - \frac{1}{4}\right)\)

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

Thus, the exact value is \(\frac{1}{4} - \frac{1}{2} e^{-1}\).

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

Applying integration by parts:

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

\(\int \frac{\ln x}{\sqrt{x}} \, dx = 2\sqrt{x} \ln x \bigg|_{1}^{4} - \int 2\sqrt{x} \cdot \frac{1}{x} \, dx\)

\(= 2\sqrt{x} \ln x \bigg|_{1}^{4} - 2 \int x^{-1/2} \, dx\)

Integrate \(\int x^{-1/2} \, dx\):

\(= 2\sqrt{x} \ln x \bigg|_{1}^{4} - 2 \cdot 2x^{1/2} \bigg|_{1}^{4}\)

\(= 2\sqrt{x} \ln x \bigg|_{1}^{4} - 4\sqrt{x} \bigg|_{1}^{4}\)

Evaluate at the limits:

\(= \left( 2\sqrt{4} \ln 4 - 4\sqrt{4} \right) - \left( 2\sqrt{1} \ln 1 - 4\sqrt{1} \right)\)

\(= (4 \ln 4 - 8) - (0 - 4)\)

\(= 4 \ln 4 - 8 + 4\)

\(= 4(\ln 4 - 1)\)

To solve \(\int_{2}^{4} 4x \ln x \, dx\), we use integration by parts.

Let \(u = \ln x\) and \(dv = 4x \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = 2x^2\).

Using integration by parts, \(\int u \, dv = uv - \int v \, du\), we have:

\(\int 4x \ln x \, dx = 2x^2 \ln x - \int 2x^2 \cdot \frac{1}{x} \, dx\).

Simplifying the integral, \(\int 2x^2 \cdot \frac{1}{x} \, dx = \int 2x \, dx = x^2\).

Thus, \(\int 4x \ln x \, dx = 2x^2 \ln x - x^2\).

Evaluate from 2 to 4:

\(\left[ 2x^2 \ln x - x^2 \right]_{2}^{4} = \left( 2(4)^2 \ln 4 - (4)^2 \right) - \left( 2(2)^2 \ln 2 - (2)^2 \right)\).

Calculate each part:

For \(x = 4\): \(2(16) \ln 4 - 16 = 32 \ln 4 - 16\).

For \(x = 2\): \(2(4) \ln 2 - 4 = 8 \ln 2 - 4\).

Combine results:

\((32 \ln 4 - 16) - (8 \ln 2 - 4) = 32 \ln 4 - 16 - 8 \ln 2 + 4\).

\(= 32 \ln 4 - 8 \ln 2 - 12\).

Since \(\ln 4 = 2 \ln 2\), substitute:

\(= 32(2 \ln 2) - 8 \ln 2 - 12\).

\(= 64 \ln 2 - 8 \ln 2 - 12\).

\(= 56 \ln 2 - 12\).

Thus, the given result is confirmed: \(56 \ln 2 - 12\).

(i) Differentiate \(f(x) = 3x e^{-2x}\) using the product rule:

\(f'(x) = 3e^{-2x} - 6xe^{-2x}\).

Substitute \(x = -\frac{1}{2}\):

\(f'\left(-\frac{1}{2}\right) = 3e - 6\left(-\frac{1}{2}\right)e = 6e\).

(ii) Use integration by parts for \(\int_{-\frac{1}{2}}^{0} 3x e^{-2x} \, dx\):

Let \(u = 3x\) and \(dv = e^{-2x} \, dx\).

Then \(du = 3 \, dx\) and \(v = -\frac{1}{2}e^{-2x}\).

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

\(= \left[-\frac{3}{2}xe^{-2x}\right]_{-\frac{1}{2}}^{0} + \frac{3}{2}\int_{-\frac{1}{2}}^{0} e^{-2x} \, dx\).

\(= \left[-\frac{3}{2}(0)e^{0} + \frac{3}{2}\left(-\frac{1}{2}e\right)e^{1}\right] + \frac{3}{2}\left[-\frac{1}{2}e^{-2x}\right]_{-\frac{1}{2}}^{0}\).

\(= 0 + \frac{3}{4}e - \frac{3}{4}\).

\(= -\frac{3}{4}\).

To solve \(\int_{0}^{1} (1-x)e^{-\frac{1}{2}x} \, dx\), we use integration by parts. Let \(u = 1-x\) and \(dv = e^{-\frac{1}{2}x} \, dx\).

Then \(du = -dx\) and \(v = -2e^{-\frac{1}{2}x}\).

Applying integration by parts:

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

\(= (1-x)(-2e^{-\frac{1}{2}x}) \bigg|_{0}^{1} + 2 \int_{0}^{1} e^{-\frac{1}{2}x} \, dx\)

Evaluating the first term:

\(= [-2(1-x)e^{-\frac{1}{2}x}] \bigg|_{0}^{1}\)

\(= [-2(0)e^{-\frac{1}{2}(1)} + 2(1)e^{0}]\)

\(= [0 + 2] = 2\)

Now, evaluate the second integral:

\(2 \int_{0}^{1} e^{-\frac{1}{2}x} \, dx = 2 \left[ -2e^{-\frac{1}{2}x} \right]_{0}^{1}\)

\(= 2 \left[ -2e^{-\frac{1}{2}} + 2 \right]\)

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

Combining both parts:

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

Thus, the integral evaluates to:

\(4e^{-\frac{1}{2}} - 2\)

To solve \(\int_{0}^{\pi} x^2 \sin x \, dx\), we use integration by parts. Let \(u = x^2\) and \(dv = \sin x \, dx\). Then \(du = 2x \, dx\) and \(v = -\cos x\).

Applying integration by parts:

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

\(\int x^2 \sin x \, dx = -x^2 \cos x + \int 2x \cos x \, dx\)

Now, integrate \(\int 2x \cos x \, dx\) by parts again. Let \(u = 2x\) and \(dv = \cos x \, dx\). Then \(du = 2 \, dx\) and \(v = \sin x\).

\(\int 2x \cos x \, dx = 2x \sin x - \int 2 \sin x \, dx\)

\(= 2x \sin x + 2 \cos x\)

Substituting back, we have:

\(\int x^2 \sin x \, dx = -x^2 \cos x + 2x \sin x + 2 \cos x\)

Evaluate from 0 to \(\pi\):

\(\left[ -x^2 \cos x + 2x \sin x + 2 \cos x \right]_{0}^{\pi}\)

At \(x = \pi\):

\(-\pi^2 \cos \pi + 2\pi \sin \pi + 2 \cos \pi = \pi^2 + 2(-1) = \pi^2 - 2\)

At \(x = 0\):

\(-0^2 \cos 0 + 2 \cdot 0 \cdot \sin 0 + 2 \cos 0 = 2\)

Subtracting, we get:

\((\pi^2 - 2) - 2 = \pi^2 - 4\)

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

Using integration by parts, \(\int u \, dv = uv - \int v \, du\).

Substitute the values: \(\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx\).

This simplifies to \(x \ln x - \int 1 \, dx = x \ln x - x + C\).

Evaluate from 2 to 4:

\(\left[ x \ln x - x \right]_{2}^{4} = (4 \ln 4 - 4) - (2 \ln 2 - 2)\).

Calculate each term:

\(4 \ln 4 - 4 = 4 \cdot 2 \ln 2 - 4 = 8 \ln 2 - 4\).

\(2 \ln 2 - 2 = 2 \ln 2 - 2\).

Combine the results:

\((8 \ln 2 - 4) - (2 \ln 2 - 2) = 8 \ln 2 - 4 - 2 \ln 2 + 2\).

\(= 6 \ln 2 - 2\).

To solve \(\int_{0}^{1} xe^{2x} \, dx\), we use integration by parts. Let \(u = x\) and \(dv = e^{2x} \, dx\).

Then \(du = dx\) and \(v = \frac{1}{2}e^{2x}\).

Using integration by parts, \(\int u \, dv = uv - \int v \, du\), we have:

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

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

Now, evaluate from 0 to 1:

\(\left[ \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x} \right]_{0}^{1}\).

At \(x = 1\): \(\frac{1}{2} \cdot 1 \cdot e^{2} - \frac{1}{4}e^{2} = \frac{1}{4}e^{2}\).

At \(x = 0\): \(\frac{1}{2} \cdot 0 \cdot e^{0} - \frac{1}{4}e^{0} = -\frac{1}{4}\).

Thus, the definite integral is:

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

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

Using integration by parts, \(\int u \, dv = uv - \int v \, du\), we have:

\(\int x \ln x \, dx = \frac{1}{2} x^2 \ln x - \int \frac{1}{2} x^2 \cdot \frac{1}{x} \, dx\)

\(= \frac{1}{2} x^2 \ln x - \frac{1}{2} \int x \, dx\)

\(= \frac{1}{2} x^2 \ln x - \frac{1}{2} \cdot \frac{1}{2} x^2 + C\)

\(= \frac{1}{2} x^2 \ln x - \frac{1}{4} x^2 + C\)

Now, evaluate from 1 to 2:

\(\left[ \frac{1}{2} x^2 \ln x - \frac{1}{4} x^2 \right]_{1}^{2}\)

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

\(= \left( 2 \ln 2 - 1 \right) - \left( 0 - \frac{1}{4} \right)\)

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

\(= 2 \ln 2 - \frac{3}{4}\)

(i) Let \(y = \frac{\cos x}{\sin x} = \cot x\). Differentiate using the quotient rule:

\(\frac{dy}{dx} = \frac{(\sin x)(-\sin x) - (\cos x)(\cos x)}{\sin^2 x}\)

\(= \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}\)

\(= \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x}\)

\(= \frac{-1}{\sin^2 x} = -\csc^2 x\)

(ii) Integrate by parts: Let \(u = x\) and \(dv = \csc^2 x \, dx\).

Then \(du = dx\) and \(v = -\cot x\).

\(\int x \csc^2 x \, dx = -x \cot x + \int \cot x \, dx\)

\(= -x \cot x + \ln |\sin x|\)

Evaluate from \(\frac{1}{4}\pi\) to \(\frac{1}{2}\pi\):

\(\left[ -x \cot x + \ln |\sin x| \right]_{\frac{1}{4}\pi}^{\frac{1}{2}\pi}\)

At \(x = \frac{1}{2}\pi\), \(-x \cot x = 0\) and \(\ln |\sin x| = 0\).

At \(x = \frac{1}{4}\pi\), \(-x \cot x = -\frac{1}{4}\pi\) and \(\ln |\sin x| = \ln \frac{1}{\sqrt{2}} = -\frac{1}{2} \ln 2\).

\(= 0 - \left( -\frac{1}{4}\pi - \frac{1}{2} \ln 2 \right)\)

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

\(= \frac{1}{4}(\pi + \ln 4)\)

(i) To find the \(x\)-coordinate of the maximum point \(M\), we need to find the derivative of \(y = x^2 e^{2-x}\) and set it to zero.

Using the product rule, the derivative is:

\(\frac{dy}{dx} = 2x e^{2-x} - x^2 e^{2-x}\).

Setting \(\frac{dy}{dx} = 0\), we have:

\(2x e^{2-x} - x^2 e^{2-x} = 0\).

Factor out \(e^{2-x}\):

\(e^{2-x} (2x - x^2) = 0\).

Since \(e^{2-x} \neq 0\), we solve \(2x - x^2 = 0\).

\(x(2 - x) = 0\).

Thus, \(x = 0\) or \(x = 2\). Since \(M\) is a maximum point, \(x = 2\).

(ii) To find \(\int_0^2 x^2 e^{2-x} \, dx\), use integration by parts twice.

First integration by parts:

Let \(u = x^2\), \(dv = e^{2-x} \, dx\).

Then \(du = 2x \, dx\), \(v = -e^{2-x}\).

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

Second integration by parts on \(\int 2x e^{2-x} \, dx\):

Let \(u = 2x\), \(dv = e^{2-x} \, dx\).

Then \(du = 2 \, dx\), \(v = -e^{2-x}\).

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

\(\int 2 e^{2-x} \, dx = -2 e^{2-x}\).

Thus, \(\int x^2 e^{2-x} \, dx = -x^2 e^{2-x} - 2x e^{2-x} - 2 e^{2-x}\).

Evaluate from 0 to 2:

\(\left[ -x^2 e^{2-x} - 2x e^{2-x} - 2 e^{2-x} \right]_0^2\).

At \(x = 2\), the expression is \(-4e^0 - 4e^0 - 2e^0 = -10\).

At \(x = 0\), the expression is \(0\).

Therefore, the integral is \(2e^2 - 10\).

(i) To find \(\frac{dy}{dx}\), use the product rule: \(\frac{d}{dx}(uv) = u'v + uv'\).

Let \(u = x\) and \(v = \cos \frac{1}{2}x\). Then \(u' = 1\) and \(v' = -\frac{1}{2} \sin \frac{1}{2}x\).

Thus, \(\frac{dy}{dx} = 1 \cdot \cos \frac{1}{2}x + x \cdot \left(-\frac{1}{2} \sin \frac{1}{2}x\right) = \cos \frac{1}{2}x - \frac{1}{2}x \sin \frac{1}{2}x\).

To find \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx}\) again using the product rule.

\(\frac{d^2y}{dx^2} = -\frac{1}{2} \sin \frac{1}{2}x - \frac{1}{2} \left( \cos \frac{1}{2}x + \frac{1}{2}x \sin \frac{1}{2}x \right)\).

Simplify to get \(\frac{d^2y}{dx^2} = -\frac{1}{2} \sin \frac{1}{2}x - \frac{1}{4}x \cos \frac{1}{2}x - \frac{1}{2} \sin \frac{1}{2}x\).

Substitute into \(4 \frac{d^2y}{dx^2} + y + 4 \sin \frac{1}{2}x = 0\) to verify.

(ii) To find the area, integrate \(y = x \cos \frac{1}{2}x\) from \(0\) to \(\pi\).

\(\int_0^\pi x \cos \frac{1}{2}x \, dx = \left[ 2x \sin \frac{1}{2}x - 4 \cos \frac{1}{2}x \right]_0^\pi\).

Evaluate: \(2\pi - 4\).

(i) To find the minimum point \(M\), we first find the derivative of \(y = x^2 \ln x\) using the product rule:

\(\frac{dy}{dx} = 2x \ln x + x\).

Set the derivative to zero to find critical points:

\(2x \ln x + x = 0\).

\(x(2 \ln x + 1) = 0\).

Since \(x = 0\) is not in the domain, solve \(2 \ln x + 1 = 0\):

\(\ln x = -\frac{1}{2}\).

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

Substitute back to find \(y\):

\(y = (e^{-\frac{1}{2}})^2 \ln(e^{-\frac{1}{2}}) = e^{-1}(-\frac{1}{2}) = -\frac{1}{2}e^{-1}\).

Thus, \(M \left( e^{-\frac{1}{2}}, -\frac{1}{2}e^{-1} \right)\).

(ii) To find the area of the shaded region, integrate \(y = x^2 \ln x\) from \(x = 1\) to \(x = e\):

Use integration by parts:

Let \(u = \ln x\), \(dv = x^2 dx\).

Then \(du = \frac{1}{x} dx\), \(v = \frac{x^3}{3}\).

\(\int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx\).

\(= \frac{x^3}{3} \ln x - \frac{1}{3} \int x^2 \, dx\).

\(= \frac{x^3}{3} \ln x - \frac{1}{9} x^3\).

Evaluate from \(x = 1\) to \(x = e\):

\(\left[ \frac{x^3}{3} \ln x - \frac{1}{9} x^3 \right]_1^e\).

\(= \left( \frac{e^3}{3} \cdot 1 - \frac{1}{9} e^3 \right) - \left( \frac{1}{3} \cdot 0 - \frac{1}{9} \right)\).

\(= \frac{e^3}{3} - \frac{1}{9} e^3 + \frac{1}{9}\).

\(= \frac{1}{9}(2e^3 + 1)\).

(i) To find the area under the curve \(y = x^2 e^{-x}\) from \(x = 0\) to \(x = 3\), we integrate:

\(\int_0^3 x^2 e^{-x} \, dx\)

Using integration by parts, let \(u = x^2\) and \(dv = e^{-x} \, dx\). Then \(du = 2x \, dx\) and \(v = -e^{-x}\).

\(\int x^2 e^{-x} \, dx = -x^2 e^{-x} + \int 2x e^{-x} \, dx\)

Apply integration by parts again to \(\int 2x e^{-x} \, dx\):

Let \(u = 2x\) and \(dv = e^{-x} \, dx\). Then \(du = 2 \, dx\) and \(v = -e^{-x}\).

\(\int 2x e^{-x} \, dx = -2x e^{-x} + \int 2 e^{-x} \, dx\)

\(= -2x e^{-x} - 2e^{-x}\)

Substitute back:

\(\int x^2 e^{-x} \, dx = -x^2 e^{-x} - 2x e^{-x} - 2e^{-x}\)

Evaluate from \(x = 0\) to \(x = 3\):

\(\left[ -x^2 e^{-x} - 2x e^{-x} - 2e^{-x} \right]_0^3\)

At \(x = 3\):

\(-9e^{-3} - 6e^{-3} - 2e^{-3} = -17e^{-3}\)

At \(x = 0\):

\(0 - 0 - 2 = -2\)

Thus, the area is:

\(-(-2) - (-17e^{-3}) = 2 - \frac{17}{e^3}\)

(ii) To find the maximum point, differentiate \(y = x^2 e^{-x}\):

\(\frac{dy}{dx} = 2x e^{-x} - x^2 e^{-x} = e^{-x}(2x - x^2)\)

Set \(\frac{dy}{dx} = 0\):

\(e^{-x}(2x - x^2) = 0\)

\(2x - x^2 = 0\)

\(x(2 - x) = 0\)

\(x = 0\) or \(x = 2\). Since \(x = 0\) is not a maximum, \(x = 2\) is the maximum point.

(iii) The tangent at \(P\) passes through the origin. The equation of the tangent at \(x = a\) is:

\(y - a^2 e^{-a} = (2a - a^2)e^{-a}(x - a)\)

Substitute \((0,0)\):

\(0 - a^2 e^{-a} = (2a - a^2)e^{-a}(0 - a)\)

\(a^2 = a^2(2 - a)\)

\(2 - a = 1\)

\(a = 1\). Thus, \(x = 1\).

(i) To find the coordinates of \(M\), we first find the derivative of \(y = x^3 \ln x\) using the product rule:

\(\frac{dy}{dx} = 3x^2 \ln x + x^2\).

Set the derivative to zero to find the critical points:

\(3x^2 \ln x + x^2 = 0\).

Factor out \(x^2\):

\(x^2 (3 \ln x + 1) = 0\).

Since \(x \neq 0\), solve \(3 \ln x + 1 = 0\):

\(\ln x = -\frac{1}{3}\).

\(x = e^{-1/3}\).

Substitute \(x = e^{-1/3}\) back into the original equation to find \(y\):

\(y = (e^{-1/3})^3 \ln(e^{-1/3}) = -\frac{1}{3e}\).

Thus, the coordinates of \(M\) are \(\left( e^{-1/3}, -\frac{1}{3e} \right)\).

(ii) To find the area of the shaded region, integrate \(y = x^3 \ln x\) from \(x = 1\) to \(x = 2\):

\(\int_1^2 x^3 \ln x \, dx\).

Using integration by parts, let \(u = \ln x\) and \(dv = x^3 \, dx\):

\(du = \frac{1}{x} \, dx\), \(v = \frac{x^4}{4}\).

\(\int x^3 \ln x \, dx = \frac{x^4}{4} \ln x - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx\).

\(= \frac{x^4}{4} \ln x - \frac{1}{4} \int x^3 \, dx\).

\(= \frac{x^4}{4} \ln x - \frac{1}{16} x^4 + C\).

Evaluate from \(x = 1\) to \(x = 2\):

\(\left[ \frac{x^4}{4} \ln x - \frac{1}{16} x^4 \right]_1^2\).

\(= \left( \frac{16}{4} \ln 2 - \frac{1}{16} \cdot 16 \right) - \left( \frac{1}{4} \ln 1 - \frac{1}{16} \right)\).

\(= 4 \ln 2 - 1 - 0 + \frac{1}{16}\).

\(= 4 \ln 2 - \frac{15}{16}\).

(i) The curve cuts the \(x\)-axis where \(y = 0\). For \(y = \frac{\ln x}{\sqrt{x}} = 0\), \(\ln x = 0\) which gives \(x = 1\). Thus, the coordinates of \(A\) are \((1, 0)\).

(ii) To find the maximum point \(M\), we differentiate \(y = \frac{\ln x}{\sqrt{x}}\) using the quotient rule:

\(\frac{d}{dx} \left( \frac{\ln x}{\sqrt{x}} \right) = \frac{\sqrt{x} \cdot \frac{1}{x} - \ln x \cdot \frac{1}{2}x^{-1/2}}{x} = \frac{1 - \frac{1}{2} \ln x}{x^{3/2}}\).

Setting the derivative to zero gives:

\(1 - \frac{1}{2} \ln x = 0\)

\(\ln x = 2\)

\(x = e^2\).

(iii) To find the area under the curve from \(x = 1\) to \(x = 4\), we use integration by parts:

Let \(u = \ln x\) and \(dv = \frac{1}{\sqrt{x}} \, dx\), then \(du = \frac{1}{x} \, dx\) and \(v = 2\sqrt{x}\).

\(\int \frac{\ln x}{\sqrt{x}} \, dx = 2\sqrt{x} \ln x - \int 2\sqrt{x} \cdot \frac{1}{x} \, dx\)

\(= 2\sqrt{x} \ln x - 2 \int x^{-1/2} \, dx\)

\(= 2\sqrt{x} \ln x - 4\sqrt{x}\).

Evaluating from \(x = 1\) to \(x = 4\):

\(\left[ 2\sqrt{x} \ln x - 4\sqrt{x} \right]_1^4 = (4 \ln 4 - 8) - (0 - 4)\)

\(= 4 \ln 4 - 4\)

\(= 8 \ln 2 - 4\).

(i) To find the \(x\)-coordinate of the minimum point \(M\), we first find the derivative of \(y = x^{\frac{1}{2}} \ln x\) using the product rule:

\(\frac{d}{dx}(x^{\frac{1}{2}} \ln x) = \frac{1}{2}x^{-\frac{1}{2}} \ln x + x^{\frac{1}{2}} \cdot \frac{1}{x} = \frac{1}{2}x^{-\frac{1}{2}} \ln x + x^{-\frac{1}{2}}\).

Set the derivative to zero to find critical points:

\(\frac{1}{2}x^{-\frac{1}{2}} \ln x + x^{-\frac{1}{2}} = 0\).

\(\ln x = -2 \Rightarrow x = e^{-2}\).

(ii) To find the area of the shaded region, use integration by parts on \(\int x^{\frac{1}{2}} \ln x \, dx\) from \(x = 1\) to \(x = 4\).

Let \(u = \ln x\) and \(dv = x^{\frac{1}{2}} \, dx\), then \(du = \frac{1}{x} \, dx\) and \(v = \frac{2}{3}x^{\frac{3}{2}}\).

Integration by parts gives:

\(\int x^{\frac{1}{2}} \ln x \, dx = \frac{2}{3}x^{\frac{3}{2}} \ln x - \int \frac{2}{3}x^{\frac{3}{2}} \cdot \frac{1}{x} \, dx\).

\(= \frac{2}{3}x^{\frac{3}{2}} \ln x - \frac{2}{3} \int x^{\frac{1}{2}} \, dx\).

\(= \frac{2}{3}x^{\frac{3}{2}} \ln x - \frac{2}{3} \cdot \frac{2}{3}x^{\frac{3}{2}} + C\).

Evaluate from \(x = 1\) to \(x = 4\):

\(\left[ \frac{2}{3}x^{\frac{3}{2}} \ln x - \frac{4}{9}x^{\frac{3}{2}} \right]_1^4\).

Calculate the definite integral:

\(\left( \frac{2}{3} \cdot 8 \cdot \ln 4 - \frac{4}{9} \cdot 8 \right) - \left( 0 - \frac{4}{9} \right) = 4.28\).

(i) To find the \(x\)-coordinate of \(M\), we need to find the derivative of \(y = x^2 e^{-\frac{1}{2}x}\) and set it to zero. Using the product rule, the derivative is:

\(\frac{d}{dx}(x^2 e^{-\frac{1}{2}x}) = 2x e^{-\frac{1}{2}x} - \frac{1}{2}x^2 e^{-\frac{1}{2}x}\).

Setting the derivative to zero:

\(2x e^{-\frac{1}{2}x} - \frac{1}{2}x^2 e^{-\frac{1}{2}x} = 0\).

Factor out \(e^{-\frac{1}{2}x}\):

\(e^{-\frac{1}{2}x} (2x - \frac{1}{2}x^2) = 0\).

Since \(e^{-\frac{1}{2}x} \neq 0\), solve \(2x - \frac{1}{2}x^2 = 0\):

\(x(2 - \frac{1}{2}x) = 0\).

\(x = 0\) or \(x = 4\).

Since \(x = 0\) is not a maximum, \(x = 4\) is the maximum point.

(ii) To find the area of the shaded region, integrate \(y = x^2 e^{-\frac{1}{2}x}\) from \(x = 0\) to \(x = 1\):

Use integration by parts: \(\int x^2 e^{-\frac{1}{2}x} \, dx\).

Let \(u = x^2\), \(dv = e^{-\frac{1}{2}x} \, dx\).

Then \(du = 2x \, dx\), \(v = -2e^{-\frac{1}{2}x}\).

\(\int x^2 e^{-\frac{1}{2}x} \, dx = -2x^2 e^{-\frac{1}{2}x} + 4 \int x e^{-\frac{1}{2}x} \, dx\).

Integrate \(\int x e^{-\frac{1}{2}x} \, dx\) by parts again.

Let \(u = x\), \(dv = e^{-\frac{1}{2}x} \, dx\).

Then \(du = dx\), \(v = -2e^{-\frac{1}{2}x}\).

\(\int x e^{-\frac{1}{2}x} \, dx = -2x e^{-\frac{1}{2}x} + 4 \int e^{-\frac{1}{2}x} \, dx\).

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

Complete the integration:

\(-2(x^2 + 4x + 8)e^{-\frac{1}{2}x}\).

Evaluate from \(x = 0\) to \(x = 1\):

\([-2(1^2 + 4 \cdot 1 + 8)e^{-\frac{1}{2}}] - [-2(0^2 + 4 \cdot 0 + 8)e^{0}]\).

\(= -26e^{-\frac{1}{2}} + 16\).

Final area: \(16 - 26e^{-\frac{1}{2}}\).

(i) To find the \(x\)-coordinate of \(A\), set \(y = 0\) in the equation \(y = \frac{\ln x}{x^2}\). This gives \(\ln x = 0\), so \(x = 1\).

(ii) To find the coordinates of \(M\), first find the derivative of \(y = \frac{\ln x}{x^2}\) using the quotient rule: \(y' = \frac{2 \ln x}{x^3} + \frac{1}{x^2}\). Set \(y' = 0\) to find the critical points: \(\frac{2 \ln x}{x^3} + \frac{1}{x^2} = 0\). Solving for \(\ln x\), we get \(x = e^{\frac{1}{2}}\). Substitute back to find \(y\): \(y = \frac{1}{2e}\). Thus, the coordinates of \(M\) are \(\left( e^{\frac{1}{2}}, \frac{1}{2e} \right)\).

(iii) To find the area, use integration by parts on \(\int_1^e \frac{\ln x}{x^2} \, dx\). Let \(u = \ln x\) and \(dv = \frac{1}{x^2} \, dx\), then \(du = \frac{1}{x} \, dx\) and \(v = -\frac{1}{x}\). The integration by parts formula gives:

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

Evaluate the integral: \(-\frac{\ln x}{x} - \frac{1}{x}\) from 1 to \(e\). This results in:

\(\left[ -\frac{\ln x}{x} - \frac{1}{x} \right]_1^e = \left( -\frac{1}{e} - \frac{1}{e} \right) - \left( 0 - 1 \right) = 1 - \frac{2}{e}\).

Answer: (i) \(x=\dfrac{7}{2}\). (ii) The area is \(\dfrac14(5+e^{-6})\).

(i) The curve is

\(y=(3-x)e^{-2x}.\)

Differentiate using the product rule:

\(\frac{dy}{dx}=(-1)e^{-2x}+(3-x)(-2e^{-2x}).\)

So

\(\frac{dy}{dx}=(-1-6+2x)e^{-2x}=(2x-7)e^{-2x}.\)

At the stationary point,

\((2x-7)e^{-2x}=0.\)

Since \(e^{-2x}\ne0\),

\(2x-7=0.\)

Therefore

\(x=\frac72.\)

(ii) The curve crosses the \(x\)-axis when

\((3-x)e^{-2x}=0,\)

so \(x=3\). The required area is therefore

\(\int_0^3(3-x)e^{-2x}\,dx.\)

Using integration by parts,

\(\int(3-x)e^{-2x}\,dx=-\frac12(3-x)e^{-2x}+\frac14e^{-2x}.\)

Thus the area is

\(\left[-\frac12(3-x)e^{-2x}+\frac14e^{-2x}\right]_0^3.\)

At \(x=3\), this is \(\dfrac14e^{-6}\). At \(x=0\), this is \(-\dfrac32+\dfrac14=-\dfrac54\). Hence the area is

\(\frac14e^{-6}-\left(-\frac54\right)=\frac14(5+e^{-6}).\)

(i) To find the \(x\)-coordinate of \(M\), we need to find the derivative of \(y = (x + 1) e^{-\frac{1}{3}x}\) and set it to zero to find the critical points.

Using the product rule, the derivative is:

\(\frac{dy}{dx} = -\frac{1}{3}(x + 1)e^{-\frac{1}{3}x} + e^{-\frac{1}{3}x}\)

Setting \(\frac{dy}{dx} = 0\), we solve for \(x\):

\(-\frac{1}{3}(x + 1) + 1 = 0\)

\(-\frac{1}{3}x - \frac{1}{3} + 1 = 0\)

\(-\frac{1}{3}x + \frac{2}{3} = 0\)

\(x = 2\)

(ii) To find the area of the shaded region, we integrate \(y = (x + 1) e^{-\frac{1}{3}x}\) from \(x = -1\) to \(x = 0\).

Using integration by parts, we have:

\(\int (x + 1) e^{-\frac{1}{3}x} \, dx = -3(x + 1)e^{-\frac{1}{3}x} - 9e^{-\frac{1}{3}x}\)

Evaluating from \(x = -1\) to \(x = 0\):

\(\left[-3(x + 1)e^{-\frac{1}{3}x} - 9e^{-\frac{1}{3}x}\right]_{-1}^{0}\)

\(= \left[-3(0 + 1)e^{0} - 9e^{0}\right] - \left[-3(-1 + 1)e^{\frac{1}{3}} - 9e^{\frac{1}{3}}\right]\)

\(= [-3 - 9] - [0 - 9e^{-\frac{1}{3}}]\)

\(= -12 + 9e^{-\frac{1}{3}}\)

\(= 9e^{-3} - 12\)

(i) To find the stationary points, we need to differentiate \(y = (1 + x^2) e^{-\frac{1}{2}x}\) and set the derivative to zero. Using the product rule, let \(u = 1 + x^2\) and \(v = e^{-\frac{1}{2}x}\). Then \(\frac{du}{dx} = 2x\) and \(\frac{dv}{dx} = -\frac{1}{2} e^{-\frac{1}{2}x}\).

The derivative is \(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = (1 + x^2)(-\frac{1}{2} e^{-\frac{1}{2}x}) + e^{-\frac{1}{2}x}(2x)\).

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

Setting \(\frac{dy}{dx} = 0\), we solve \(-\frac{1}{2} - \frac{1}{2}x^2 + 2x = 0\).

Multiplying through by \(-2\), we get \(x^2 - 4x - 1 = 0\).

Using the quadratic formula, \(x = \frac{4 \pm \sqrt{16 + 4}}{2} = 2 \pm \sqrt{3}\).

(ii) To find the area of \(R\), we integrate \(y = (1 + x^2) e^{-\frac{1}{2}x}\) from \(x = 0\) to \(x = 2\).

Using integration by parts, let \(u = 1 + x^2\) and \(dv = e^{-\frac{1}{2}x} dx\). Then \(du = 2x dx\) and \(v = -2e^{-\frac{1}{2}x}\).

The integral becomes \(\int (1 + x^2) e^{-\frac{1}{2}x} dx = -2(1 + x^2) e^{-\frac{1}{2}x} + \int 4x e^{-\frac{1}{2}x} dx\).

Integrating by parts again for \(\int 4x e^{-\frac{1}{2}x} dx\), we find \(-8xe^{-\frac{1}{2}x} - 2x^2 e^{-\frac{1}{2}x}\).

Evaluating from \(x = 0\) to \(x = 2\), we find the area is \(18 - \frac{42}{e}\).

(i) The derivative of \(y = (\ln x)^2\) is \(\frac{2 \ln x}{x}\). At \(x = e\), the gradient of the tangent is \(\frac{2 \ln e}{e} = \frac{2}{e}\). The gradient of the normal is the negative reciprocal, \(-\frac{e}{2}\). The equation of the normal at \(P(e, 1)\) is \(y - 1 = -\frac{e}{2}(x - e)\). Setting \(y = 0\) to find \(Q\), we solve \(0 - 1 = -\frac{e}{2}(x - e)\), giving \(x = e + \frac{2}{e}\).

(ii) To show \(\int \ln x \, dx = x \ln x - x + c\), use integration by parts: let \(u = \ln x\) and \(dv = dx\), then \(du = \frac{1}{x} dx\) and \(v = x\). Thus, \(\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + c\).

(iii) The area under the curve from \(x = 1\) to \(x = e\) is \(\int_1^e (\ln x)^2 \, dx\). Using integration by parts, let \(u = (\ln x)^2\) and \(dv = dx\), then \(du = \frac{2 \ln x}{x} dx\) and \(v = x\). Thus, \(\int (\ln x)^2 \, dx = x(\ln x)^2 - \int x \cdot \frac{2 \ln x}{x} \, dx = x(\ln x)^2 - 2 \int \ln x \, dx\). From part (ii), \(\int \ln x \, dx = x \ln x - x\), so \(\int (\ln x)^2 \, dx = x(\ln x)^2 - 2(x \ln x - x) = x(\ln x)^2 - 2x \ln x + 2x\). Evaluating from \(x = 1\) to \(x = e\), we find the area is \(e - 2\). The area of the triangle formed by the normal is \(\frac{1}{2} \times \frac{2}{e} \times 1 = \frac{1}{e}\). The total area is \(e - 2 + \frac{1}{e}\).

(i) To find the \(x\)-coordinate of the maximum point \(M\), we need to find the derivative of \(y = (2x - x^2)e^{\frac{1}{2}x}\) and set it to zero.

Using the product rule, the derivative is:

\(\frac{d}{dx}[(2x - x^2)e^{\frac{1}{2}x}] = (2 - 2x)e^{\frac{1}{2}x} + \frac{1}{2}(2x - x^2)e^{\frac{1}{2}x}\)

Setting the derivative to zero:

\((2 - 2x)e^{\frac{1}{2}x} + \frac{1}{2}(2x - x^2)e^{\frac{1}{2}x} = 0\)

Simplifying and solving for \(x\), we find:

\(x = \sqrt{5} - 1\)

(ii) To find the area of the shaded region, we need to integrate \(y = (2x - x^2)e^{\frac{1}{2}x}\) from \(x = 0\) to \(x = 2\).

Using integration by parts, we have:

\(\int (2x - x^2)e^{\frac{1}{2}x} \, dx\)

Integrating by parts twice, we obtain:

\((12x - 2x^2 - 24)e^{\frac{1}{2}x}\)

Evaluating from \(x = 0\) to \(x = 2\):

\([12(2) - 2(2)^2 - 24]e^{1} - [12(0) - 2(0)^2 - 24]e^{0}\)

\(= 24e - 8e - 24\)

\(= 24 - 8e\)

(a) Let \(u = \\sin x\), then \(du = \\cos x \, dx\). The integral becomes \(\int \\sin 2x \\cos^2 x \, dx = \int 2u(1-u^2) \, du\).

Integrate to get \(\int 2(u - u^3) \, du = 2 \left( \frac{u^2}{2} - \frac{u^4}{4} \right) = u^2 - \frac{u^4}{2}\).

Evaluate from \(u = 0\) to \(u = 1\):

\(\left[ u^2 - \frac{u^4}{2} \right]_0^1 = \left( 1 - \frac{1}{2} \right) - (0 - 0) = \frac{1}{2}\).

(b) Differentiate \(y = \\sin 2x \\cos^2 x\) using the product rule:

\(\frac{dy}{dx} = 2 \\cos 2x \\cos^2 x - 2 \\sin 2x \\cos x \\sin x\).

Set \(\frac{dy}{dx} = 0\) and solve:

\(2 \\cos 2x \\cos^2 x = 2 \\sin 2x \\cos x \\sin x\).

Using double angle formulas, simplify to \(4 \\sin^2 x = 1\), giving \(\\sin^2 x = \frac{1}{4}\).

Thus, \(\\sin x = \frac{1}{2}\), so \(x = \frac{1}{6}\pi\).

(i) To find the \(x\)-coordinate of \(M\), we need to find the maximum of \(y = e^{\cos x} \sin^3 x\). Differentiate using the product rule and chain rule:

\(\frac{dy}{dx} = e^{\cos x} \left( -\sin x \sin^3 x + 3\cos x \sin^2 x \right)\).

Set \(\frac{dy}{dx} = 0\) to find critical points:

\(-\sin x \sin^3 x + 3\cos x \sin^2 x = 0\).

Factor out \(\sin^2 x\):

\(\sin^2 x (-\sin x + 3\cos x) = 0\).

Thus, \(\sin x = 0\) or \(-\sin x + 3\cos x = 0\).

For \(-\sin x + 3\cos x = 0\), solve for \(\cos^2 x + 3\cos x - 1 = 0\).

Using the quadratic formula, find \(\cos x\) and then \(x\):

\(x = 1.26\) (to 2 decimal places).

(ii) Use the substitution \(u = \cos x\), then \(du = -\sin x \, dx\).

The integral becomes \(\int e^u (u^2 - 1) \, du\).

Integrate by parts:

\(\int e^u (u^2 - 1) \, du = e^u (u^2 - 1) - 2\int ue^u \, du\).

Complete the integration to obtain:

\(e^u (u^2 - 2u + 1)\).

Substitute back the limits \(u = 1\) and \(u = -1\) (or \(x = 0\) and \(x = \pi\)):

The area \(R\) is \(\frac{4}{e}\).

(i) Let \(x = t^2 + 1\). Then \(dx = 2t \, dt\).

When \(t = 0\), \(x = 0^2 + 1 = 1\).

When \(t = 2\), \(x = 2^2 + 1 = 5\).

Substitute into the integral:

\(I = \int_0^2 4t^3 \ln(t^2 + 1) \, dt = \int_1^5 (2x - 2) \ln x \, dx\).

(ii) Use integration by parts:

Let \(u = \ln x\) and \(dv = (2x - 2) \, dx\).

Then \(du = \frac{1}{x} \, dx\) and \(v = x^2 - 2x\).

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

\(= (x^2 - 2x) \ln x - \int (x - 2) \, dx\).

\(= (x^2 - 2x) \ln x - \left( \frac{1}{2}x^2 - 2x \right)\).

Evaluate from 1 to 5:

\(= [(5^2 - 2 \cdot 5) \ln 5 - \left( \frac{1}{2} \cdot 5^2 - 2 \cdot 5 \right)] - [(1^2 - 2 \cdot 1) \ln 1 - \left( \frac{1}{2} \cdot 1^2 - 2 \cdot 1 \right)]\).

\(= [15 \ln 5 - (\frac{25}{2} - 10)] - [0 - (\frac{1}{2} - 2)]\).

\(= 15 \ln 5 - \frac{5}{2} + \frac{3}{2}\).

\(= 15 \ln 5 - 4\).

(i) The minimum point \(A\) occurs where \(x = 1\) because \((\ln x)^2\) is minimized when \(\ln x = 0\), i.e., \(x = 1\).

(ii) First, find \(f'(x) = 2 \frac{\ln x}{x}\). Then, differentiate again to find \(f''(x) = 2 \left( \frac{1}{x} \cdot \frac{-1}{x^2} + \frac{\ln x}{x^2} \right) = 2 \left( \frac{-1}{x^2} + \frac{\ln x}{x^2} \right)\). At \(x = e\), \(f''(e) = 2 \left( \frac{-1}{e^2} + \frac{1}{e^2} \right) = 0\).

(iii) The area is given by \(\int_1^e (\ln x)^2 \, dx\). Using the substitution \(x = e^u\), \(dx = e^u \, du\), and the limits change from \(x = 1\) to \(u = 0\) and \(x = e\) to \(u = 1\). The integral becomes \(\int_0^1 u^2 e^u \, du\).

(iv) Evaluate \(\int_0^1 u^2 e^u \, du\) using integration by parts. Let \(v = u^2\) and \(dw = e^u \, du\), then \(dv = 2u \, du\) and \(w = e^u\). The integral becomes \(\left[ u^2 e^u \right]_0^1 - \int_0^1 2u e^u \, du\). Apply integration by parts again to \(\int_0^1 2u e^u \, du\) with \(v = u\) and \(dw = 2e^u \, du\). The result is \(e - 2\).