The Fundamental Theorem and accumulated change
Введено| English | Значение |
|---|---|
| accumulation function/əˌkjuːmjʊˈleɪʃn ˈfʌŋkʃn/ | функция накопления |
How does the accumulated volume recover that same rate when differentiated?
- A flow meter reports a changing rate.
- This lesson studies accumulation function 累积函数: A function whose value is the signed integral from a fixed starting point to a variable endpoint.
The method
- For continuous f on an interval, A(x)=integral from a to x of f(t) dt has A′(x)=f(x).
- If F′=f, the definite integral from a to b is F(b)−F(a).
- These two directions connect accumulation and differentiation.
- Dummy variable t is integrated; x is the moving endpoint.
- For A(g(x)), apply the chain rule: its derivative is f(g(x))g′(x).
For A(x)=∫₁ˣ(3t²-2)dt, why are A(2)=5 and A′(2)=10 different numbers?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The Fundamental Theorem has two directions: accumulation differentiates back to the integrand, A′(x)=f(x), while the definite integral F(b)-F(a) measures total change. Here x³-2x+1 at 2 is 5, but f(2)=10.
A worked case
Let A(x)=integral from 1 to x of (3t²−2) dt. An antiderivative is t³−2t, so A(x)=x³−2x+1, A(1)=0, A(2)=5 and A′(2)=10. Thus accumulated change 5 differs from endpoint rate 10. For B(x)=integral from 1 to x² of (3t²−2) dt, B′(x)=(3x⁴−2)2x; at x=1 this is 2. Independently, if F′=3x²−2 and F(1)=4, then F=x³−2x+5. Its change F(2)−F(1)=5 agrees with A(2), although its initial value is not zero.
The Fundamental Theorem and accumulated change
For continuous f on an interval, A(x)=integral from a to x of f(t) dt has A′(x)=f(x)
Check the hypothesis, endpoints and coefficients behind each integral calculation.
Find A(2) for the worked accumulation function.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
A(2)=8−4+1=5.
Where it goes wrong
- Subtract the lower antiderivative value even when the lower limit is not zero.
- A definite integral is signed change, not automatically total positive area.
- A′(x) is the integrand at the endpoint, not its antiderivative.
- Continuity on the relevant interval is a sufficient hypothesis; do not integrate across a pole such as 1/t at zero.
- A moving endpoint x² also contributes its chain-rule factor.
Not true: The derivative of an accumulation function is its accumulated total. The derivative of ∫ₐˣ f(t) dt is f(x), the integrand's value at x, not the total.
Find A′(2).
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
A′(2)=3×4−2=10.
The derivative of an accumulation function is its accumulated total.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The derivative of ∫ₐˣ f(t) dt is f(x), the integrand's value at x, not the total.
Applying it
- Use rate units multiplied by input units for accumulated change.
- Check A(a)=0 and differentiate the final expression to recover f.
- Initial values determine constants in a state function; definite changes do not depend on an arbitrary antiderivative constant.
- This lesson gives the theorem and elementary variable-bound examples; improper integrals are outside this treatment.
Find B′(1) for the worked moving endpoint.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
B′(1)=(3−2)×2=2.
Match each theorem component to its statement.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The dummy variable t is integrated out; the endpoint x is what the answer depends on, and a non-linear endpoint brings its own chain factor.
In your course
- AQA A-level Mathematics (7357), specification section H.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Accumulation function: A function whose value is the signed integral from a fixed starting point to a variable endpoint.
- For continuous f on an interval, A(x)=integral from a to x of f(t) dt has A′(x)=f(x).
- Watch for: subtract the lower antiderivative value even when the lower limit is not zero.