\int_0^{0.2} 30 \cdot \frac{s^5}{5} ds = 6 \int_0^{0.2} s^5 ds = 6 \cdot \frac{s^6}{6} \Big|_0^{0.2} = (0.2)^6 = \left(\frac{1}{5}\right)^6 = \frac{1}{15625}

["Understanding a Definite Integral: Step-by-Step Calculation from 0 to 0.2", "Integrals are fundamental tools in calculus with applications across physics, engineering, economics, and beyond. One intriguing integral often examined for learning purposes is:", "[\n\int_0^{0.2} 30 \cdot \frac{s^5}{5} , ds = 6 \int_0^{0.2} s^5 , ds = 6 \cdot \frac{s^6}{6} \Big|_0^{0.2} = (0.2)^6 = \left(\frac{1}{5}\right)^6 = \frac{1}{15625}\n]", "In this article, we’ll carefully unpack how this transformation simplifies integration, clarify each step, and highlight the key mathematical principles involved.", "---", "### Step 1: Simplify the Integrand", "Start with the original expression:", "[\n\int_0^{0.2} 30 \cdot \frac{s^5}{5} , ds\n]", "Notice that (30 \cdot \frac{1}{5} = 6), so the integrand simplifies neatly:", "[\n\int_0^{0.2} 6 s^5 , ds\n]", "This simplification reduces complexity and reveals a more efficient path to evaluation.", "---", "### Step 2: Factor Out Constants", "We can pull the constant (6) out of the integral:", "[\n6 \int_0^{0.2} s^5 , ds\n]", "Integrating constants is straightforward, and this manipulation highlights the basic integration rule:", "[\n\int s^n , ds = \frac{s^{n+1}}{n+1} + C\n]", "For (n = 5), this yields (\frac{s^6}{6}).", "---", "### Step 3: Apply the Limits of Integration", "Now compute the definite integral:", "[\n6 \int_0^{0.2} s^5 , ds = 6 \left[ \frac{s^6}{6} \right]_0^{0.2} = 6 \cdot \left( \frac{(0.2)^6}{6} - \frac{0^6}{6} \right)\n]", "Since (0^6 = 0), we get:", "[\n6 \cdot \frac{(0.2)^6}{6} = (0.2)^6\n]", "---", "### Step 4: Evaluate the Final Expression", "Now compute ( (0.2)^6 ):", "First, express (0.2) as a fraction:", "[\n0.2 = \frac{1}{5}\n]", "So:", "[\n(0.2)^6 = \left(\frac{1}{5}\right)^6 = \frac{1}{5^6} = \frac{1}{15625}\n]", "---", "### Why This Breakdown Matters", "- Simplifying integrands improves computational efficiency and reduces error risk.\n- Factoring constants demonstrates a core integration algebra rule.\n- Recognizing a (s^5) antiderivative shows direct application of the power rule.\n- Evaluating at bounds ends the problem cleanly.\n- Converting decimal to fraction gives exact form and emphasizes symbolic clarity.", "---", "### Final Result", "[\n\boxed{ \int_0^{0.2} 30 \cdot \frac{s^5}{5} , ds = (0.2)^6 = \frac{1}{15625} }\n]", "This integral is a concise illustration of how algebraic manipulation, integration rules, and number conversion combine to solve and simplify real-world problems in calculus.", "---", "Related Topics:\n- Power rule for integration\n- Evaluating definite integrals\n- Fractions and exponents in algebra\n- Real-world applications of definite integrals", "Keywords: integral calculus, definite integral, power function integration, ∫₀⁰·² s⁵ ds, (1/5)⁶, simplifying integrals, mathematical computation, calculus fundamentals."]