Use integration by parts with \( u = 4x \), \( dv = e^{2x}dx \).

Then \( du = 4dx \), \( v = \tfrac{1}{2}e^{2x} \).

\[ \int_0^1 4x e^{2x}dx = \Bigl[2xe^{2x}\Bigr]_0^1 - \int_0^1 2e^{2x}dx = \Bigl[2xe^{2x} - e^{2x}\Bigr]_0^1 = (2e^2 - e^2) - (0 - 1) = 1 + e^2. \]

Numerically, \( 1 + e^2 \approx 8.39 \) (3 s.f.).

Answer: 8.39.

Take \( u = 3x \), \( dv = \sin 2x\,dx \).

Then \( du = 3dx \), \( v = -\tfrac{1}{2}\cos 2x \).

\[ \int_0^{\pi/2} 3x\sin 2x\,dx = \Bigl[-\tfrac{3}{2}x\cos 2x\Bigr]_0^{\pi/2} + \int_0^{\pi/2} \tfrac{3}{2}\cos 2x\,dx = \Bigl[-\tfrac{3}{2}x\cos 2x + \tfrac{3}{4}\sin 2x\Bigr]_0^{\pi/2}. \]

At \( x = \tfrac{\pi}{2} \): value = \( -\tfrac{3}{2}\tfrac{\pi}{2}\cos \pi + 0 = \tfrac{3\pi}{4} \).

At \( x = 0 \): value = 0.

So the exact answer is \( \dfrac{3\pi}{4} \approx 2.36 \) (3 s.f.).

Answer: 2.36.

Write \( (2x+1)e^x = (2x-1)e^x + 2e^x \).

By parts (or by inspection), \( \int (2x-1)e^x dx = (2x-1)e^x + C \).

So

\[ \int_0^2 (2x+1)e^x dx = \Bigl[(2x-1)e^x\Bigr]_0^2 + \Bigl[2e^x\Bigr]_0^2 = (3e^2 - (-1)) + (2e^2 - 2) = 3e^2 + 1. \]

Numerically, \( 3e^2 + 1 \approx 23.2 \) (3 s.f.).

Answer: 23.2.

Let \( u = x \), \( dv = \cos(\tfrac{x}{2})dx \).

Then \( du = dx \), \( v = 2\sin(\tfrac{x}{2}) \).

\[ \int_0^{\pi/2} x\cos\Bigl(\tfrac{x}{2}\Bigr)dx = \Bigl[2x\sin\tfrac{x}{2}\Bigr]_0^{\pi/2} - \int_0^{\pi/2} 2\sin\tfrac{x}{2}\,dx. \]

Next, \( \int 2\sin(\tfrac{x}{2})dx = -4\cos(\tfrac{x}{2}) + C \).

So

\[ = \Bigl[2x\sin\tfrac{x}{2}\Bigr]_0^{\pi/2} - \Bigl[-4\cos\tfrac{x}{2}\Bigr]_0^{\pi/2} = \frac{\pi}{\sqrt{2}} - \bigl(-\tfrac{4}{\sqrt{2}} + 4\bigr) = \frac{\pi+4}{\sqrt{2}} - 4. \]

Numerically this is \( \approx 1.05 \) (3 s.f.).

Answer: 1.05.

Take \( u = \ln x \), \( dv = x^4 dx \).

Then \( du = \tfrac{1}{x}dx \), \( v = \tfrac{1}{5}x^5 \).

\[ \int_1^{e} x^4\ln x\,dx = \Bigl[\tfrac{1}{5}x^5 \ln x\Bigr]_1^{e} - \int_1^{e} \tfrac{1}{5}x^5\cdot\tfrac{1}{x}dx = \Bigl[\tfrac{1}{5}x^5 \ln x\Bigr]_1^{e} - \int_1^{e} \tfrac{1}{5}x^4 dx. \]

\[ = \Bigl[\tfrac{1}{5}x^5 \ln x\Bigr]_1^{e} - \Bigl[\tfrac{1}{25}x^5\Bigr]_1^{e} = \Bigl(\tfrac{1}{5}e^5 - 0\Bigr) - \Bigl(\tfrac{1}{25}e^5 - \tfrac{1}{25}\Bigr) = \tfrac{4}{25}e^5 + \tfrac{1}{25}. \]

Numerically this is \( \approx 23.8 \) (3 s.f.).

Answer: 23.8.