So ∫₁² (0.5)^u du = [ -u / ln(2) ]₁² = ( -2/ln2 ) - ( -1/ln2 ) = (-2 + 1)/ln2 = -1/ln2

["Understanding the Definite Integral: So ∫₁² (0.5)^u du = –u / ln(2) from 1 to 2 = –1/ln(2)", "Mathematics is filled with elegant expressions where seemingly complex problems simplify beautifully through integration. One such example involves computing the definite integral of an exponential function. This article explores the precise evaluation of the integral ∫₁² (0.5)^u du, demonstrating how fundamental principles of calculus yield a clear, concise result — and revealing its elegant closed-form expression in terms of natural logarithms.", "---", "### What is the Integral of (0.5)^u from u = 1 to u = 2?", "The integral\n∫₁² (0.5)^u du\nrepresents the net area under the curve of the exponential decay function (0.5)^u between u = 1 and u = 2. Since (0.5)^u = e^{u · ln(0.5)} and ln(0.5) = –ln(2), this function describes a continuous decrease, vital in modeling phenomena like radioactive decay, cooling processes, and computational efficiency decay.", "---", "### Step-by-Step Evaluation", "We begin by computing the indefinite integral:", "[\n\int (0.5)^u , du = \int e^{u \ln(0.5)} , du = \int e^{-u \ln(2)} , du\n]", "Using the standard formula ∫e^{au} du = (1/a)e^{au} + C, with a = –ln(2):", "[\n\int (0.5)^u , du = -\frac{1}{\ln 2} \cdot (0.5)^u + C\n]", "Now, evaluate the definite integral from 1 to 2:", "[\n\int_1^2 (0.5)^u , du = \left[ -\frac{1}{\ln 2} \cdot (0.5)^u \right]_1^2\n]", "Substitute the upper and lower limits:", "[\n= -\frac{1}{\ln 2} (0.5)^2 - \left( -\frac{1}{\ln 2} (0.5)^1 \right)\n= -\frac{1}{\ln 2} \left( (0.5)^2 - (0.5)^1 \right)\n]", "Simplify the powers:", "[\n(0.5)^2 = 0.25, \quad (0.5)^1 = 0.5\n]", "So:", "[\n= -\frac{1}{\ln 2} (0.25 - 0.5) = -\frac{1}{\ln 2} (-0.25) = \frac{0.25}{\ln 2} = \frac{1}{4 \ln 2}\n]", "Wait — but this contradicts the expression –u / ln(2) evaluated from 1 to 2 given in the problem. Let’s re-express everything carefully.", "---", "### Re-evaluating Using Natural Logarithm Form", "We recall:\n(0.5)^u = e^{-u ln 2}\n∫ (0.5)^u du = –(1/ln 2) e^{-u ln 2} + C", "So definite integral:", "[\n\int_1^2 (0.5)^u du = \left[ -\frac{1}{\ln 2} (0.5)^u \right]_1^2\n= -\frac{1}{\ln 2} \left( (0.5)^2 - (0.5)^1 \right)\n= -\frac{1}{\ln 2} \left( \frac{1}{4} - \frac{1}{2} \right)\n= -\frac{1}{\ln 2} \left( -\frac{1}{4} \right)\n= \frac{1}{4 \ln 2}\n]", "However, note the problem statement claims the result is –u / ln(2) evaluated from 1 to 2, which yields:", "[\n\left[ -\frac{u}{\ln 2} \right]_1^2 = \left( -\frac{2}{\ln 2} \right) - \left( -\frac{1}{\ln 2} \right) = \frac{-2 + 1}{\ln 2} = \frac{-1}{\ln 2}\n]", "So there’s a mismatch unless the integral was misinterpreted.", "---", "### Resolving the Discrepancy", "The confusion arises because (0.5)^u = 2^{-u}, and its antiderivative is –(1/ln 2) · 2^{-u}, not –u / ln 2. However, the expression –u / ln 2 appears only if we integrate by parts or Mistakenly assume a linear antiderivative.", "But let’s recompute carefully:\nIf we mistakenly differentiate –u/ln(2), we get –1/ln(2), which is the rate, not the area.", "Thus, the correct evaluation of ∫ (0.5)^u du = –(1/ln 2)·2^{-u} + C", "Then:", "∫₁² (0.5)^u du = –(1/ln 2)[2^{-2} – 2^{-1}]\n= –(1/ln 2)[0.25 – 0.5]\n= –(1/ln 2)(–0.25)\n= 0.25 / ln 2 = 1/(4 ln 2)", "This confirms the previously derived value.", "---", "### Why Does the Problem Mention –u / ln(2) Between 1 and 2?", "This suggests a misunderstanding — integrating (0.5)^u does not yield a result involving u linearly in the antiderivative. The integral ∫ₐᵇ b^u du equals\n[\n\frac{b^u}{\ln b}, \quad b > 0, b <br/>\ne 1\n]\nor in exponential form,\n[\n\frac{1}{\ln b}(b^u)\big|_a^b = \frac{b^u}{\ln b}(b - a)\n]", "So for b = 0.5:", "∫₁² (0.5)^u du = –(1/ln 2) (0.5² – 0.5¹) = –(1/ln 2)(0.25 – 0.5) = (0.25)/ln 2", "---", "### Correct Final Expression", "Thus, the accurate value is:", "[\n\int_1^2 (0.5)^u du = \frac{1}{4 \ln 2}\n]", "But if we strictly interpret the problem’s claim —\n[\n- \int_1^2 \frac{u}{\ln 2} du = - \frac{1}{\ln 2} \int_1^2 u,du = - \frac{1}{\ln 2} \left[ \frac{u^2}{2} \right]_1^2\n= - \frac{1}{\ln 2} \left( \frac{4}{2} - \frac{1}{2} \right) = - \frac{1}{\ln 2} \cdot \frac{3}{2} = -\frac{3}{2 \ln 2}\n]", "This differs entirely — so the original integral cannot equal –u/ln(2) from 1 to 2.", "---", "### Clarifying the True Value and Counterfactual", "Despite this, the core insight remains invaluable:\nThe integral of an exponential function (even non-integer base like 0.5) yields a logarithmic expression, not a linear one.\nThus, the expression –u/ln(2) integrated from 1 to 2 is not correct, but it serves as an educational springboard for understanding:", "[\n\int a^u du = -\frac{a^u}{\ln a} + C, \quad a > 0, a <br/>\ne 1\n]", "Evaluating from 1 to 2:", "[\n\left[ -\frac{(0.5)^u}{\ln 0.5} \right]_1^2 = \left[ \frac{(0.5)^u}{\ln 2} \right]_1^2 = \frac{0.25}{\ln 2} - \frac{0.5}{-\ln 2} = \frac{0.25 + 0.5}{\ln 2} = \frac{0.75}{\ln 2} = \frac{3}{4 \ln 2}\n]", "This confirms again that the correct numerator is magnitude 1/4, not 3/4.", "---", "### Summary: What to Remember", "- The integral ∫₁² (0.5)^u du evaluates exactly to –(1/ln 2) ( (0.5)^2 – (0.5)^1 ) = –(1/ln 2)( –0.25 ) = 0.25 / ln 2 = 1/(4 ln 2)\n- The expression –u/ln(2) integrated from 1 to 2 gives –1/ln 2 (2 – 1) = –1/ln 2, which does not match — this likely stems from confusing linear integration with exponential decay.\n- The correct form –a^u / ln a ensures logarithmic decay, essential for understanding decay processes, algorithm complexity, and probability distributions.", "---", "### Final Thought", "While the expression involving –u / ln 2 per u may misrepresent the integral, the deeper lesson endures: integration of exponential functions unveils logarithmic relationships — a cornerstone of calculus with profound implications across science, engineering, and applied mathematics.", "Whether computing signal decay, modeling half-life, or analyzing recursive algorithms, mastering such integrals empowers precise quantitative reasoning in a complex world.", "---", "Further Reading:\n- Integration of exponential functions\n- Natural logarithm and calculus\n- Applications of definite integrals in decay models", "---\nKeywords: ∫₁² (0.5)^u du, (0.5)^u integral, natural logarithm, exponential decay integration, calculus examples, logarithmic integral, definite integral meaning."]