Let's solve the given problems:

Problem (a):

$\int_0^2 xe^{3x} \, dx$

Solution:

We use integration by parts: $\int u \, dv = uv - \int v \, du$

Let:

• $u = x$, so $du = dx$
• $dv = e^{3x} dx$, so $v = \frac{e^{3x}}{3}$

Now substitute: $\int xe^{3x} dx = \left[ \frac{x e^{3x}}{3} \right]_0^2 - \int_0^2 \frac{e^{3x}}{3} dx$

1. Evaluate $\left[ \frac{x e^{3x}}{3} \right]_0^2$: $\frac{2e^{6}}{3} - \frac{0e^0}{3} = \frac{2e^{6}}{3}$

2. Solve $\int_0^2 \frac{e^{3x}}{3} dx$: $\frac{1}{3} \int_0^2 e^{3x} dx = \frac{1}{3} \left[ \frac{e^{3x}}{3} \right]_0^2 = \frac{1}{9} \left( e^6 - 1 \right)$

Now combine: $\int_0^2 xe^{3x} dx = \frac{2e^6}{3} - \frac{1}{9} (e^6 - 1)$

Simplify: $\int_0^2 xe^{3x} dx = \frac{6e^6}{9} - \frac{e^6}{9} + \frac{1}{9} = \frac{5e^6 + 1}{9}$

Problem (b):

$\int_1^{e^4} \frac{\ln x}{x^7} \, dx$

Solution:

Let $u = \ln x$, so $du = \frac{1}{x} dx$. Also, $x^7 = e^{7u}$, and the limits transform:

• When $x = 1$, $u = \ln 1 = 0$
• When $x = e^4$, $u = \ln e^4 = 4$

Substituting: $\int_1^{e^4} \frac{\ln x}{x^7} dx = \int_0^4 \frac{u}{e^{7u}} \, du$

Simplify: $\int_0^4 u e^{-7u} \, du$

We use integration by parts:

• Let $u = u$, so $du = du$
• Let $dv = e^{-7u} du$, so $v = \frac{e^{-7u}}{-7}$

Now substitute: $\int u e^{-7u} du = \left[ \frac{u e^{-7u}}{-7} \right]_0^4 - \int_0^4 \frac{e^{-7u}}{-7} du$

1. Evaluate $\left[ \frac{u e^{-7u}}{-7} \right]_0^4$: $\frac{4 e^{-28}}{-7} - \frac{0 e^0}{-7} = -\frac{4 e^{-28}}{7}$

2. Solve $\int_0^4 \frac{e^{-7u}}{-7} du$: $-\frac{1}{7} \int_0^4 e^{-7u} du = -\frac{1}{7} \left[ \frac{e^{-7u}}{-7} \right]_0^4 = \frac{1}{49} \left( e^{-28} - 1 \right)$

Now combine: $\int_0^4 u e^{-7u} du = -\frac{4 e^{-28}}{7} - \frac{1}{49} (e^{-28} - 1)$

Simplify: $\int_0^4 u e^{-7u} du = -\frac{28 e^{-28}}{49} - \frac{e^{-28} - 1}{49}$

$\int_0^4 u e^{-7u} du = \frac{-28 e^{-28} - e^{-28} + 1}{49} = \frac{-29 e^{-28} + 1}{49}$

Final Answers:

(a) $\frac{5e^6 + 1}{9}$

(b) $\frac{-29 e^{-28} + 1}{49}$

Would you like a step-by-step breakdown of any of these parts? Here are related questions for further exploration:

1. How does integration by parts apply in different contexts?
2. How can substitution simplify integrals with logarithms?
3. How do limits change during substitution in definite integrals?
4. Can you generalize these solutions for similar exponential or logarithmic forms?
5. How would these integrals change if the bounds were infinite?

Tip: For problems involving exponential functions, always consider substitution or integration by parts as the first approach.