Substitution, parts and partial fractions · Changement de variable, intégration par parties et fractions partielles
| English | Français |
|---|---|
| integration by parts/ˌɪntɪˈɡreɪʃn baɪ pɑːts/ | intégration par parties |
Can we integrate the two factors separately?
- A simple-looking product such as xe^x cannot be integrated by separately integrating each factor.
- This lesson studies integration by parts 分部积分法: An integration method based on the derivative of a product.
Choose the mathematical structure
- Use substitution when an inner derivative appears as a factor. By parts, integral u v prime =uv-integral u prime v. For a rational function, divide first if needed, then decompose into partial fractions.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines integration by parts? · Quelle description définit correctement l'intégration par parties ?
An integration method based on the derivative of a product. · Une méthode d'intégration basée sur la dérivée d'un produit.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
For integral xe^x dx, take u=x and v prime=e^x. Then the integral is xe^x-e^x+C. For integral 2x/(x²+1) dx, substitute w=x²+1, dw=2x dx; the answer is ln(x²+1)+C.
Substitution, parts and partial fractions · Changement de variable, intégration par parties et fractions partielles
V/π = integral from 0 to 3 of x² dx = 9 · V/π = intégrale de 0 à 3 de x² dx = 9
The widget shows the volume integrand y²=x² for y=x; compare the exact value 9 with the volume check. · Le widget affiche l'intégrande du volume y²=x² pour y=x ; comparez la valeur exacte 9 avec le contrôle du volume.
Evaluate the integral of xe^x from 0 to 1. · Évaluez l'intégrale de xe^x de 0 à 1.
An antiderivative is (x-1)e^x. At 1 and 0 it gives 0 and -1, so the integral is 1. · Une primitive est (x-1)e^x. En 1 et 0, elle donne 0 et -1, donc l'intégrale est 1.
Test a tempting shortcut
- Choosing u and v prime well matters: the remaining integral should become simpler. In a definite substitution, either change the limits or return to x before applying the original limits.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
The integral of a product equals the product of its separate integrals. This claim is false. Explain which definition or assumption it violates.
Find the coefficient of ln(x²+1) in the integral of 2x/(x²+1). · Trouvez le coefficient de ln(x²+1) dans l'intégrale de 2x/(x²+1).
Substitution w=x²+1 gives dw=2x dx, so the coefficient of ln(x²+1) is 1. · Le changement de variable w=x²+1 donne dw=2x dx, donc le coefficient de ln(x²+1) est 1.
The integral of a product equals the product of its separate integrals. · L'intégrale d'un produit n'est pas égale au produit de ses intégrales séparées.
Choosing u and v prime well matters: the remaining integral should become simpler. In a definite substitution, either change the limits or return to x before applying the original limits. · Bien choisir u et v prime est essentiel : l'intégrale restante doit devenir plus simple. Pour un changement de variable défini, modifiez les bornes ou revenez à x avant d'appliquer les bornes originales.
Interpret a new situation
- Check by differentiating. For volumes of revolution around the x-axis use V=π integral y² dx; do not confuse the square of a function with the integral of the function.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find the volume divided by π when y=x is revolved about the x-axis from 0 to 3. · Trouvez le volume divisé par π lorsque y=x est tourné autour de l'axe des x de 0 à 3.
V/π=integral x² dx from 0 to 3=[x³/3]=9. · V/π=intégrale de x² dx de 0 à 3=[x³/3]=9.
Match each part of a complete solution to its purpose. · Associer chaque partie d'une solution complète à son but.
An assumption justifies the model; a check tests the result; interpretation connects it to the question. · Une hypothèse justifie le modèle ; une vérification teste le résultat ; l'interprétation le relie à la question.
Use this in your course
- edexcel IAL pure mathematics; official unit FP3. Other-unit enrichment is identified in the scope review; it is not an extra cash-in requirement.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
An integration method based on the derivative of a product. Choose the relationship, show the method, check its assumptions and interpret the result.