\( \displaystyle \int e^x\,dx = e^x + C \).

So \( \displaystyle \int_0^{\ln 2} e^x\,dx = [e^x]_0^{\ln 2} = e^{\ln 2} - e^0 = 2 - 1 = 1 \).

Answer: 1.

\( \displaystyle \int e^x\,dx = e^x + C \).

\( \displaystyle \int_0^{2\ln 3} e^x\,dx = [e^x]_0^{2\ln 3} = e^{2\ln 3} - e^0 = 3^2 - 1 = 9 - 1 = 8 \).

Answer: 8.

\( \displaystyle \int_{\ln 2}^{\ln 3} e^x\,dx = [e^x]_{\ln 2}^{\ln 3} = e^{\ln 3} - e^{\ln 2} = 3 - 2 = 1 \).

Answer: 1.

\( \displaystyle \int 4e^{4x}\,dx = e^{4x} + C \).

So \( \displaystyle \int_0^{\ln 3} 4e^{4x}\,dx = [e^{4x}]_0^{\ln 3} = e^{4\ln 3} - e^0 = 3^4 - 1 = 81 - 1 = 80 \).

Answer: 80.

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

So \( \displaystyle \int_0^{\ln 4} 2e^{2x}\,dx = [e^{2x}]_0^{\ln 4} = e^{2\ln 4} - e^0 = 4^2 - 1 = 16 - 1 = 15 \).

Answer: 15.

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

So \( \displaystyle \int_0^{\ln 5} e^{-x}\,dx = [-e^{-x}]_0^{\ln 5} = -e^{-\ln 5} - (-e^0) = -\tfrac{1}{5} + 1 = \tfrac{4}{5}. \)

\( \tfrac{4}{5} = 0.8 \) (3 s.f.).

Answer: 0.8.

Note that \( e^{x\ln 3} = 3^x \).

Let \( u = x\ln 3 \), so \( du = \ln 3\,dx \), \( dx = \dfrac{1}{\ln 3}du \).

Then

\[ \int_0^1 e^{x\ln 3}\,dx = \frac{1}{\ln 3} \int_{0}^{\ln 3} e^u\,du = \frac{1}{\ln 3}[e^u]_0^{\ln 3} = \frac{1}{\ln 3}(3 - 1) = \frac{2}{\ln 3}. \]

Numerically, \( \dfrac{2}{\ln 3} \approx 1.82 \) (3 s.f.).

Answer: 1.82.

Note that \( e^{x\ln 2} = 2^x \).

Let \( u = x\ln 2 \), so \( du = \ln 2\,dx \), \( dx = \dfrac{1}{\ln 2}du \).

Then

\[ \int_{-2}^{1} e^{x\ln 2}\,dx = \frac{1}{\ln 2}[e^{x\ln 2}]_{-2}^{1} = \frac{1}{\ln 2}(2^1 - 2^{-2}) = \frac{1}{\ln 2}\Bigl(2 - \frac{1}{4}\Bigr) = \frac{7}{4\ln 2}. \]

Numerically, \( \dfrac{7}{4\ln 2} \approx 2.52 \) (3 s.f.).

Answer: 2.52.

Note that \( e^{-x\ln 3} = 3^{-x} \).

Let \( u = -x\ln 3 \), so \( du = -\ln 3\,dx \), \( dx = -\dfrac{1}{\ln 3}du \).

Alternatively, integrate directly:

\( \displaystyle \int e^{-x\ln 3}\,dx = -\frac{1}{\ln 3}e^{-x\ln 3} + C \).

So

\[ \int_{-1}^{4} e^{-x\ln 3}\,dx = \Bigl[-\frac{1}{\ln 3}e^{-x\ln 3}\Bigr]_{-1}^{4} = -\frac{1}{\ln 3}\Bigl(3^{-4} - 3^{1}\Bigr) = \frac{1}{\ln 3}\Bigl(3 - \frac{1}{81}\Bigr). \]

\( 3 - \tfrac{1}{81} = \tfrac{242}{81} \), so the exact value is \( \dfrac{242}{81\,\ln 3} \).

Numerically, this is approximately \( 2.72 \) (3 s.f.).

Answer: 2.72.

\( \displaystyle \int e^x\,dx = e^x + C \).

So \( \displaystyle \int_0^{3} e^x\,dx = [e^x]_0^{3} = e^3 - 1 \).

Numerically, \( e^3 - 1 \approx 19.1 \) (3 s.f.).

Answer: 19.1.

\( \displaystyle \int e^{-2x}\,dx = -\tfrac{1}{2}e^{-2x} + C \).

So \( \displaystyle \int_0^{2} e^{-2x}\,dx = \Bigl[-\tfrac{1}{2}e^{-2x}\Bigr]_0^{2} = -\tfrac{1}{2}e^{-4} - \Bigl(-\tfrac{1}{2}\Bigr) = \tfrac{1}{2}(1 - e^{-4}) \).

Numerically, \( \tfrac{1}{2}(1 - e^{-4}) \approx 0.491 \) (3 s.f.).

Answer: 0.491.

Integrate term by term:

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

So

\[ \int_0^{2} (e^x - e^{-x})\,dx = [e^x + e^{-x}]_0^{2} = (e^2 + e^{-2}) - (1 + 1) = e^2 + e^{-2} - 2. \]

Numerically, \( e^2 + e^{-2} - 2 \approx 5.52 \) (3 s.f.).

Answer: 5.52.