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.

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

Then \( du = dx \), \( v = -\cos x \).

\[ \int_0^{\pi/2} x\sin x\,dx = \Bigl[-x\cos x\Bigr]_0^{\pi/2} + \int_0^{\pi/2} \cos x\,dx = 0 + [\sin x]_0^{\pi/2} = 1. \]

Answer (3 s.f.): 1.00.

Use integration by parts twice. One convenient route:

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

Then integrate \( 4x e^x \) by parts again to get

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

Substitute limits:

\[ (9e^2 - 1) - (8e^2 - 0) + (4e^2 - 4) = 5e^2 - 5. \]

Numerically, \( 5e^2 - 5 \approx 31.9 \) (3 s.f.).

Answer: 31.9.

Integrate by parts twice:

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

Thus

\[ = (e - 0) - (2e - 0) + (2e - 2) = e - 2. \]

Numerically, \( e - 2 \approx 0.718 \) (3 s.f.).

Answer: 0.718.

First integration by parts:

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

Then integrate \( 2x\cos x \) by parts:

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

Putting it all together gives

\[ \int_0^{\pi/2} x^2\sin x\,dx = \pi - 2. \]

Numerically, \( \pi - 2 \approx 1.14 \) (3 s.f.).

Answer: 1.14.

Integrate by parts, letting \( u = x^2 \), \( dv = \sin 2x\,dx \).

After two integrations by parts, one obtains

\[ \int_{\pi/6}^{\pi/3} x^2 \sin 2x \, dx = \frac{5\pi^2}{144} + \frac{\pi\sqrt{3}}{24} - \frac{1}{4}. \]

Numerically, this is approximately \( 0.319 \) (3 s.f.).

Answer: 0.319.

Consider \( \displaystyle \int_{-A}^{0} x^2 e^{3x}dx \) and let \( A \to \infty \).

Repeated integration by parts yields

\[ \int x^2 e^{3x} dx = \tfrac{1}{3}x^2 e^{3x} - \tfrac{2}{9}x e^{3x} + \tfrac{2}{27}e^{3x} + C. \]

Evaluating from \( -\infty \) to \( 0 \), all terms at \( -\infty \) vanish, leaving

\[ \int_{-\infty}^{0} x^2 e^{3x} dx = \tfrac{2}{27}. \]

Numerically, \( \tfrac{2}{27} \approx 0.0741 \) (3 s.f.).

Answer: 0.0741.

Integrate by parts twice (or use the standard form for \( \int e^{ax}\cos bx\,dx \)).

After using parts and solving for the original integral, one obtains an exact expression of the form

\[ \int_0^{\pi/4} e^{3x} \cos 2x \, dx = \frac{1}{13}\Bigl(2e^{3\pi/4} - 3\Bigr). \]

Numerically, this is approximately \( 1.39 \) (3 s.f.).

Answer: 1.39.

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

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

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

So

\[ = (-e^{-4} - 0) + (-\tfrac{1}{4}e^{-4} + \tfrac{1}{4}) = \tfrac{1}{4} - \tfrac{5}{4}e^{-4}. \]

Numerically, this is \( \approx 0.227 \) (3 s.f.).

Answer: 0.227.

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

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

\[ \int_0^{\pi/6} x\cos 2x\,dx = \Bigl[\tfrac{1}{2}x\sin 2x\Bigr]_0^{\pi/6} - \int_0^{\pi/6} \tfrac{1}{2}\sin 2x\,dx. \]

Now \( \int \tfrac{1}{2}\sin 2x\,dx = -\tfrac{1}{4}\cos 2x + C \).

So

\[ = \Bigl[\tfrac{1}{2}x\sin 2x\Bigr]_0^{\pi/6} - \Bigl[-\tfrac{1}{4}\cos 2x\Bigr]_0^{\pi/6}. \]

Evaluating gives the exact value

\[ \frac{\pi\sqrt{3}}{24} - \frac{1}{8}. \]

Numerically, this is approximately \( 0.102 \) (3 s.f.).

Answer: 0.102.