Here is the solution for each integral:

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

To solve this, use integration by parts:

Let $u = x$ and $dv = e^{3x} dx$, so that: $du = dx, \quad v = \frac{1}{3}e^{3x}.$

Using the integration by parts formula: $\int u \, dv = uv - \int v \, du,$ we have: $\int xe^{3x} \, dx = \left( x \cdot \frac{1}{3}e^{3x} \right) - \int \left( \frac{1}{3}e^{3x} \cdot 1 \right) dx.$

The second term becomes: $\int \frac{1}{3}e^{3x} dx = \frac{1}{9}e^{3x}.$

So, the solution to the integral is: $\int xe^{3x} dx = \frac{x}{3}e^{3x} - \frac{1}{9}e^{3x} + C.$

Now evaluate it from $x = 0$ to $x = 2$: [ \int_0^2 xe^{3x} , dx = \left[ \frac{2}{3}e^{6} - \frac{1}{9}e^{6} \right] - \left[ \frac{0}{3}e^{0} - \frac{1}{9}e^{0} \right]. ]

Simplify: $\int_0^2 xe^{3x} \, dx = \left( \frac{6}{9}e^6 - \frac{1}{9}e^6 \right) - \left( 0 - \frac{1}{9} \right).$

$\int_0^2 xe^{3x} \, dx = \frac{5}{9}e^6 + \frac{1}{9}.$

(b) $\int_1^4 \frac{\ln x}{x^7} \, dx$

Let $u = \ln x$, so that $du = \frac{1}{x} dx$. Rewrite the integral as: $\int_1^4 \frac{\ln x}{x^7} \, dx = \int_1^4 u \cdot x^{-6} dx.$

Here, $x^{-6}$ becomes part of the substitution. Using substitution $x = e^t$, simplify to get:

...

from sympy import symbols, integrate, exp, log, Rational

Define variables

x = symbols('x')

Part (a) Integral

integral_a = integrate(x * exp(3*x), (x, 0, 2))

Part (b) Integral

integral_b = integrate(log(x) / x**7, (x, 1, 4))

integral_a, integral_b(1/9 + 5exp(6)/9, 455/16384 - log(4)/24576)(1/9 + 5exp(6)/9, 455/16384 - log(4)/24576)success### Solutions:

(a) $\int_0^2 xe^{3x} \, dx = \frac{1}{9} + \frac{5}{9}e^6$

(b) $\int_1^4 \frac{\ln x}{x^7} \, dx = \frac{455}{16384} - \frac{\ln(4)}{24576}$

Would you like further clarification on any part of this process or a more detailed explanation? Here are some related questions you might find useful:

1. What is the process of integration by parts, and when do we use it?
2. How can substitutions simplify integrals involving logarithmic functions?
3. Can numerical methods be used to approximate these integrals?
4. Why does $\ln x$ require substitution in the second problem?
5. How can we interpret these integrals graphically?

Tip: Always double-check limits of integration after substitution to ensure accuracy.