The evaluation part says the integral from a to b equals F(b) − F(a), for any antiderivative F.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
The constant cancels
Any antiderivative works, because a + C on both terms subtracts away. That is why definite integrals carry no + C.
The accumulation part
If g(x) is the integral of f from a to x, then g′(x) = f(x). Differentiating an accumulation returns the integrand.
What it means
Differentiation and integration are inverse operations. That is the single most important statement in the course.
With a variable upper limit
If the upper limit is a function of x, multiply by its derivative. The Chain Rule applies here too.
Why it matters
It defines functions no formula reaches. The error function is defined exactly this way, as an accumulation.
The first part
If F(x) is the accumulated integral of f from a to x, then F′(x) = f(x). Differentiating an accumulation returns the integrand, which is far from obvious and is the heart of the theorem.
The second part
The definite integral from a to b is F(b) − F(a) for any antiderivative F. This turns an infinite limiting process into two evaluations and a subtraction.
With a variable upper limit and a chain rule
If the upper limit is g(x) rather than x, the derivative is f(g(x))·g′(x). That extra factor is the chain rule, and forgetting it is the standard error in this exam-favourite question type.
The net change theorem
The integral of a rate of change over an interval is the net change in the quantity. Stated that way, the fundamental theorem becomes the tool used in almost every applied free-response question.
Step 2: Try It Yourself
Tap and try it out.
- Point(0, 0)
- Second point(2, 4)
- Slope between them2
Slide the second point towards the first. The slope between them approaches the slope of the curve at that point.
Step 3: Watch an Example
One step at a time.
Watch Sana Differentiate an Accumulation
Sana has g(x) equal to the integral of t² from 1 to x, and needs g′(x).
- Step 1
The accumulation part says differentiating undoes the integration.
Step 4: Your Turn
Practice makes it stick.
The Evaluation
Problem 1 of 2
Integral of 3x² from 1 to 2, with antiderivative x³. What is the value?
The Accumulation
Problem 2 of 2
g(x) is the integral of t² from 1 to x. What is g′(3)?
Both Parts
1 of 8
Integral of 2x from 0 to 4, antiderivative x². Value?
2 of 8
Integral of 3x² from 0 to 2, antiderivative x³. Value?
3 of 8
g(x) is the integral of t³ from 0 to x. What is g′(2)?
4 of 8
g(x) is the integral of sin t from 0 to x. What is g′(0)?
5 of 8
Integral from 5 to 5 of any function. Value?
6 of 8
An integral from 2 to 7 is 12. What is it from 7 to 2?
7 of 8
Match each part of the theorem with what it does.
Tap a card on the left to start.
8 of 8
Integral of 2x from 1 to 3, antiderivative x². Value?
Step 5: Quick Check
Show what you know.
Question 1 of 2
g(x) is the integral of t² from 1 to x. What is g′(4)?
Question 2 of 2
What does the Fundamental Theorem establish?
What You Learned
- The evaluation part computes an integral as F(b) − F(a).
- The accumulation part says differentiating an integral returns the integrand.
- A variable upper limit brings the Chain Rule with it.