Let us evaluate the given integrals step by step.

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(a)

$I = \int_{0}^{5} x e^{4x} \, dx$

To solve this integral, we will use integration by parts:

$\int u \, dv = uv - \int v \, du$

Let:

• $u = x$ $\implies$ $du = dx$
• $dv = e^{4x} dx$ $\implies$ $v = \frac{e^{4x}}{4}$

Substitute into the formula: $I = \left[ \frac{x e^{4x}}{4} \right]_{0}^{5} - \int_{0}^{5} \frac{e^{4x}}{4} dx$

Now compute the second term: $\int \frac{e^{4x}}{4} dx = \frac{1}{4} \cdot \frac{e^{4x}}{4} = \frac{e^{4x}}{16}$

Substitute: [ I = \left[ \frac{x e^{4x}}{4} \right]{0}^{5} - \left[ \frac{e^{4x}}{16} \right]{0}^{5} ]

Evaluate the limits:

1. $\left[ \frac{x e^{4x}}{4} \right]_{0}^{5} = \frac{5 e^{20}}{4} - \frac{0}{4} = \frac{5 e^{20}}{4}$
2. $\left[ \frac{e^{4x}}{16} \right]_{0}^{5} = \frac{e^{20}}{16} - \frac{e^{0}}{16} = \frac{e^{20}}{16} - \frac{1}{16}$

Simplify: $I = \frac{5 e^{20}}{4} - \left( \frac{e^{20}}{16} - \frac{1}{16} \right)$ $I = \frac{5 e^{20}}{4} - \frac{e^{20}}{16} + \frac{1}{16}$

Combine terms: $\frac{5 e^{20}}{4} = \frac{20 e^{20}}{16}, \quad \therefore I = \frac{20 e^{20} - e^{20} + 1}{16} = \frac{19 e^{20} + 1}{16}$

Thus: $\boxed{I = \frac{19 e^{20} + 1}{16}}$

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(b)

$J = \int_{1}^{e^4} \frac{\ln x}{x^3} \, dx$

Rewrite the integral: $J = \int_{1}^{e^4} \ln x \cdot x^{-3} \, dx$

Let us use integration by parts. Let:

• $u = \ln x$ $\implies$ $du = \frac{1}{x} dx$
• $dv = x^{-3} dx$ $\implies$ $v = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$

Substitute into the formula: $J = \left[ -\frac{\ln x}{2x^2} \right]_{1}^{e^4} - \int_{1}^{e^4} -\frac{1}{2x^3} dx$

Simplify: $J = \left[ -\frac{\ln x}{2x^2} \right]_{1}^{e^4} + \frac{1}{2} \int_{1}^{e^4} \frac{1}{x^3} dx$

Step 1: Compute $\int \frac{1}{x^3} dx$: $\int \frac{1}{x^3} dx = \int x^{-3} dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$

So: $\frac{1}{2} \int_{1}^{e^4} \frac{1}{x^3} dx = \frac{1}{2} \left[ -\frac{1}{2x^2} \right]_{1}^{e^4}$ $= \frac{1}{2} \left( -\frac{1}{2(e^4)^2} + \frac{1}{2(1)^2} \right)$

= \frac{-1}{4e^8} + \frac{1}{4}$$ **Step 2: Evaluate the boundary term:** $$\left[ -\frac{\ln x}{2x^2} \right]_{1}^{e^4} = -\frac{\ln(e^4)}{2(e^4)^2} + \frac{\ln(1)}{2(1)^2}$$ $$= -\frac{4}{2e^8} + 0 = -\frac{2}{e^8}$$ Combine all terms: $$J = \left( -\frac{2}{e^8} \right) + \left( \frac{-1}{4e^8} + \frac{1}{4} \right)$$ $$J = -\frac{2}{e^8} - \frac{1}{4e^8} + \frac{1}{4}$$ Combine terms: $$-\frac{2}{e^8} = -\frac{8}{4e^8}, \quad \therefore J = -\frac{8}{4e^8} - \frac{1}{4e^8} + \frac{1}{4}$$ $$J = -\frac{9}{4e^8} + \frac{1}{4}$$ Thus: $$\boxed{J = \frac{1}{4} - \frac{9}{4e^8}}$$ --- Let me know if you'd like further clarification or additional steps detailed! Here are some related questions: 1. How does integration by parts work and when is it applied? 2. What are alternative methods to solve definite integrals? 3. How can we interpret the result of these integrals geometrically? 4. Why is it important to carefully handle exponential terms in integration? 5. What is the significance of boundary evaluation in definite integrals? **Tip:** Always double-check boundary terms in definite integrals; small errors can propagate through the solution!