∫₂⁴ 2^(-d/20) dd = 20 ∫₁² 2^(-u) du = 20 [ -2^(-u) / ln2 ]₁² = 20 [ ( -1/(2^u ln2) ) ]₁² = 20 [ -1/(2^2 ln2) + 1/(2^1 ln2) ] = 20 [ -1/(4 ln2) + 1/(2 ln2) ] = 20 [ ( -0.25 + 0.5 ) / ln2 ] = 20 × (0.25 / ln2) = 5 / ln2

["# Understanding the Integral ∫₂⁴ 2^(-d/20) dd = 20 ∫₁² 2^(-u) du = 20 [ -2^(-u) / ln2 ]₁²: A Step-by-Step Explanation", "The integral ∫₂⁴ 2^(-d/20) dd might seem daunting at first glance, but with a careful breakdown of the math, we uncover a neat transformation involving exponentials and natural logarithms. This article walks you through the evaluation of this definite integral using substitution and basic logarithmic rules, revealing an elegant connection between integration and logarithmic expressions.", "---", "### Step 1: Start with the Original Integral", "We begin with:", "$$\n\int_{2}^{4} 2^{-d/20} , dd\n$$", "Recall that any exponential function of the form ( 2^{-d/20} ) can be rewritten using the identity:", "$$\n2^{-d/20} = e^{-(d/20) \ln 2} = 2^{-d/20}\n$$", "But more conveniently, we can perform a change of variables to simplify the integral.", "---", "### Step 2: Use Substitution to Change the Limits", "Let us apply the substitution:", "$$\nu = \frac{d}{20} \implies d = 20u\n$$", "Then, the differential becomes:", "$$\ndd = 20,du\n$$", "Now, change the limits accordingly:", "- When ( d = 2 ), ( u = \frac{2}{20} = \frac{1}{10} )\n- When ( d = 4 ), ( u = \frac{4}{20} = \frac{1}{5} )", "Substituting into the integral:", "$$\n\int_{2}^{4} 2^{-d/20} , dd = \int_{1/10}^{1/5} 2^{-u} \cdot 20 , du = 20 \int_{1/10}^{1/5} 2^{-u} , du\n$$", "---", "### Step 3: Fact out the Constant", "Since 20 is constant:", "$$\n= 20 \int_{1/10}^{1/5} 2^{-u} , du\n$$", "The integral of ( 2^{-u} ) is well-known:\nRecall ( 2^{-u} = e^{-u \ln 2} ), so", "$$\n\int 2^{-u} , du = \int e^{-u \ln 2} , du = -\frac{1}{\ln 2} e^{-u \ln 2} + C = -\frac{2^{-u}}{\ln 2} + C\n$$", "---", "### Step 4: Apply Definite Integration Limits", "Now compute:", "$$\n20 \int_{1/10}^{1/5} 2^{-u} , du = 20 \left[ -\frac{2^{-u}}{\ln 2} \right]<em 1_10="1/10">{1/10}^{1/5}\n= 20 \left( -\frac{2^{-1/5}}{\ln 2} + \frac{2^{-1/10}}{\ln 2} \right)\n= 20 \cdot \frac{ -2^{-1/5} + 2^{-1/10} }{\ln 2}\n$$", "---", "### Step 5: Simplify the Expression", "Note that:", "- ( 2^{-1/5} = \frac{1}{2^{1/5}} )\n- ( 2^{-1/10} = \frac{1}{2^{1/10}} )", "We rewrite:", "$$\n= \frac{20}{\ln 2} \left( 2^{-1/10} - 2^{-1/5} \right)\n= \frac{20}{\ln 2} \left( \frac{1}{2^{1/10}} - \frac{1}{2^{1/5}} \right)\n$$", "But the original solution transforms the expression algebraically:", "$$\n20 \left( -\frac{2^{-u}}{\ln 2} \right) \right)}^{1/5} = 20 \left( -\frac{2^{-1/5}}{\ln 2} + \frac{2^{-1/10}}{\ln 2} \right) = \frac{20}{\ln 2} \left( 2^{-1/10} - 2^{-1/5\n$$", "The expression can also be rewritten using the identity ( 2^{-d/20} = 2^{-u} ), confirming the factor of 20 from ( dd = 20,du ).", "---", "### Step 6: Final Simplification to Final Exact Form", "Returning to the original claim:", "$$\n= 20 \left[ -\frac{2^{-u}}{\ln 2} \right]<em 2="2">{1/10}^{1/5} = 20 \left( -\frac{2^{-1/5}}{\ln 2} + \frac{2^{-1/10}}{\ln 2} \right)\n= \frac{20}{\ln 2} \left( -2^{-1/5} + 2^{-1/10} \right)\n$$", "Now focus on simplifying the numerical coefficient:", "$$\n-2^{-1/5} + 2^{-1/10} = 2^{-1/10} - 2^{-1/5}\n$$", "Factor out ( 2^{-1/10} ):", "$$\n= 2^{-1/10} \left( 1 - 2^{-3/10} \right)\n$$", "Thus,", "$$\n\frac{20}{\ln 2} \left( 2^{-1/10} - 2^{-1/5} \right) = \frac{20}{\ln 2} \left( 2^{-1/10} - 2^{-2/10} \right)\n= \frac{20}{\ln 2} \left( 2^{-1/10} - 2^{-1/5} \right)\n$$", "Alternatively, recognizing:", "$$\n2^{-u} = \frac{1}{2^u}\n$$", "we write:", "$$\n20 \cdot \left( -\frac{1}{2^{1/5} \ln 2} + \frac{1}{2^{1/10} \ln 2} \right)\n= \frac{20}{\ln 2} \left( \frac{1}{2^{1/10}} - \frac{1}{2^{1/5}} \right)\n$$", "This matches the earlier derived expression.", "To express it neatly:", "$$\n= \frac{20}{\ln 2} \left( 2^{-1/10} - 2^{-1/5} \right)\n$$", "---", "### Step 7: Numerical Evaluation (Optional)", "To verify:", "- ( 2^{-1/10} \approx 0.9330 )\n- ( 2^{-1/5} \approx 0.8706 )", "Thus:", "$$\n2^{-1/5} - 2^{-1/10} \approx 0.8706 - 0.9330 = -0.0624\n\Rightarrow \ ext{negative} \Rightarrow \ ext{we take positive: } 2^{-1/10} - 2^{-1/5} \approx 0.0624\n$$", "Then:", "$$\n\frac{20}{\ln 2} \cdot 0.0624 \approx \frac{20}{0.6931} \cdot 0.0624 \approx 28.85 \cdot 0.0624 \approx 1.80\n$$", "Note: The exact symbolic result is ( \frac{20}{\ln 2} (2^{-1/10} - 2^{-1/5}) ), which is more precise than numerical approximation.", "---", "### Conclusion: Why This Integral Matters", "This integral arises naturally in probability, decay processes, and scaling problems where exponential decay depends linearly on a variable. The evaluation method—substitution, logarithmic integral, and substitution back—exemplifies standard calculus techniques for handling exponential functions in definite integration.", "So, remember:\nStill integrate, change limits, substitute, apply known antiderivative, and keep careful track of signs and constants.", "---", "### Key Takeaways", "- Substitution ( u = \frac{d}{20} ) simplifies the limits and reveals the structure.\n- The integral ( \int 2^{-u} du = -\frac{2^{-u}}{\ln 2} ) is key.\n- Boundary evaluation produces the final expression:\n [\n \boxed{ \int}^{4} 2^{-d/20} , dd = \frac{20}{\ln 2} \left( 2^{-1/10} - 2^{-1/5} \right) \n ]\n- This result highlights elegant connections between exponentials and logarithms.", "For further exploration, analyze similar integrals involving base ( a ), or apply this method to other exponential integrals with vertical shifts."]