Substitution, parts and partial fractions
Введено| English | Значение |
|---|---|
| integration by parts/ˌɪntɪˈɡreɪʃn baɪ pɑːts/ | An integration method derived from the product rule, transferring differentiation from one factor to another. |
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.
The method
- 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.
Why does the substitution w=x²+1 fit the integral of 2x/(x²+1)?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
With w=x²+1, dw=2x dx, and the integral becomes ∫dw/w=ln|w|+C. Substitution is powerful precisely when the inner derivative appears as a factor.
A worked case
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
V/π = integral from 0 to 3 of x² dx = 9
The widget shows the volume integrand y²=x² for y=x; compare the exact value 9 with the volume check.
Evaluate the integral of xe^x from 0 to 1.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
An antiderivative is (x-1)e^x. At 1 and 0 it gives 0 and -1, so the integral is 1.
Where it goes wrong
- 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.
Not true: The integral of a product equals the product of its separate integrals. A product needs a method such as integration by parts or substitution: ∫x·x dx=x³/3, not (x²/2)(x²/2).
Find the coefficient of ln(x²+1) in the integral of 2x/(x²+1).
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Substitution w=x²+1 gives dw=2x dx, so the coefficient of ln(x²+1) is 1.
The integral of a product equals the product of its separate integrals.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
A product needs a method such as integration by parts or substitution: ∫x·x dx=x³/3, not (x²/2)(x²/2).
Applying it
- 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.
Find the volume divided by π when y=x is revolved about the x-axis from 0 to 3.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
V/π=integral x² dx from 0 to 3=[x³/3]=9.
Match each integral to the method that fits it.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Choose the method by structure: an inner derivative suggests substitution, a product of unlike types suggests parts, and an improper rational function needs division first.
In your course
- Pearson Edexcel International A Level Mathematics, unit P4.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Integration by parts: An integration method based on the derivative of a product.
- Use substitution when an inner derivative appears as a factor.
- Watch for: choosing u and v prime well matters: the remaining integral should become simpler.
Write from a volume of 16 pi
- Revolve y equals 4 about the x-axis from 0 to 1.
- Find the volume divided by pi.
The region under y equals 4 from x equals 0 to x equals 1 is revolved about the x-axis. What is the volume divided by pi?
16 times 1 is 16.
What is the volume divided by pi?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The integral of 16 over a length of 1 is 16. Other options — 9 (the lesson volume factor); 4 (the height of the line).