Cogito
AP Calculus AB · Chapter 6 · Lesson 1
The Definite Integral as Accumulation
Adding up infinitely many thin pieces.
11 problems · about 25 minutes · AP Calculus AB 6.1, 6.6, 6.7
What this lesson teaches
The student interprets a definite integral as accumulated area and applies the Fundamental Theorem.
- A definite integral is the limit of a sum of infinitely thin slices.
- It measures signed area, so regions below the axis subtract.
- The Fundamental Theorem evaluates it by subtracting an antiderivative at the two ends.
Warm Up
Straightforward practice. Get the method working first.
4 problemsIntegrate 2x from 0 to 6.
Answer 36
Why 36.
Integrate 2x from 0 to 5.
Answer 25
Why x² from 0 to 5.
Integrate 3 from 0 to 4.
Answer 12
Why A rectangle.
Integrate 3x² from 0 to 2.
Answer 8
Why x³ from 0 to 2.
Build It Up
The same ideas with more to keep track of.
3 problemsIntegrate 2x from 3 to 3.
Answer 0
Why No width.
Select every true statement about the definite integral.
Answer Area below the axis counts as negative.; Swapping the limits flips the sign.
Why Think about what "signed" means.
Integrate 2x from 4 to 0. What is the value?
Answer -16
Why Reversed limits flip the sign.
Stretch Yourself
Mixed problems. Work out what kind of question it is before you start.
4 problemsAn antiderivative of 4x³ is x⁴. Integrate from 0 to 2.
Answer 16
Why 2⁴ − 0.
Integrate 5 from 2 to 7.
Answer 25
Why Height 5, width 5.
The Antiderivative: Integrate 2x from 0 to 4. What is the value?
Answer 16
Why 16.
Signed Area: Integrate x from −2 to 2. What is the value?
Answer 0
Why 0 — equal area above and below.