6.8
非有理関数を含む積分は、標準的な手法では評価が難しいことが多く、特に積分関数に根号が含まれる場合には困難となります。有理化置換は、そのような積分をより扱いやすい有理形式に変換することで簡略化する体系的な手法を提供します。
線質量密度が、一定の線密度定数、特性長、および棒の左端からの距離に依存する棒を…
有理関数でない積分は標準的な方法では評価が難しい。
線形質量密度が一定の線密度、特性の長さ、左側からの距離で与えられる棒を考えます。
目的は棒の質量を求めることであり、そのためにこの密度関数を棒の長さに積分する必要があります。
立方根は積分を複雑にするため、合理化の置換が役立ちます。
新しい変数 u を導入し、u を x の立方根に等しいと定義すると、式は有理形に変換されます。これから 、x は uの立方として取ることができ、微分dxはそれに応じて導かれます。積分の限界は新しい変数に合わせて調整されます。
これらの式を積分に代入すると、完全に uで書かれた方程式が得られます。仮定を行った後、積分は単純な多項式形式に簡略化されます。
この変換積分は扱いやすく、多項式の長割り算は得られる有理関数の簡略化に役立ちます。
式を uで書き換えた後、更新された極限で積分を評価すると、棒の総質量が得られます。
このようにして積分は有理化置換を用いて解かれます。
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Q1: What is a rationalizing substitution and when should you use it?
A rationalizing substitution converts integrals with non-rational functions, particularly those containing radicals, into rational forms that are easier to evaluate. When cube roots or other radicals complicate the integrand, introducing a new variable defined as that radical simplifies the expression into a polynomial or rational function suitable for standard integration techniques.
Q2: How do you set up a rationalizing substitution for an integral with cube roots?
Define a new variable u as the cube root of the original variable. Express the original variable as a power of u, then rewrite the differential dx in terms of du. Adjust the integration limits to reflect the new variable. Substitute these expressions into the integral to transform it entirely into terms of u, creating a rational function.
Q3: Why does rationalizing substitution work for integrals with radicals?
Radicals create non-rational integrands that resist standard integration methods. By substituting a new variable equal to the radical expression, you eliminate the radical and convert the integrand into a rational or polynomial form. This transformation allows you to apply algebraic techniques like polynomial long division to simplify and integrate the resulting expression.
Q4: What role does polynomial long division play in rationalizing substitution?
After substitution, the transformed integral often yields a rational function that requires simplification. Polynomial long division separates this rational function into simpler, more manageable terms that are straightforward to integrate individually. This algebraic step is essential for breaking down complex expressions into integrable components.
Q5: How do you adjust integration limits when using a rationalizing substitution?
When you introduce a new variable u, you must convert the original limits of integration to match the new variable. If the original limits are a and b for the variable x, substitute these values into the relationship between u and x to find the new limits. This ensures the definite integral evaluates over the correct region in the transformed variable.
Q6: Can you apply rationalizing substitution to find physical quantities like mass?
Yes. For a rod with linear mass density involving radicals, rationalizing substitution transforms the density function into an integrable form. After substitution and simplification, evaluating the transformed integral with updated limits yields the total mass. This demonstrates how rationalizing substitution solves real-world integration problems involving non-rational functions.
Q7: How does rationalizing substitution relate to other integration techniques?
Rationalizing substitution converts non-rational integrands into rational forms, which can then be handled using integration of rational functions using partial fractions or other algebraic methods. It serves as a preprocessing step that transforms difficult integrals into standard forms amenable to established integration techniques.