Cogito
Calculus · Chapter 8 · Lesson 2
Evaluating Definite Integrals
Antiderivative at the top, minus at the bottom.
12 problems · about 22 minutes · FUN-6.A, FUN-6.B
What this lesson teaches
The student evaluates definite integrals using the Fundamental Theorem and interprets negative area.
- Evaluate a definite integral as F(b) − F(a).
- The constant of integration cancels, so no + C is needed.
- Area below the axis counts as negative, so an integral is signed area.
Warm Up
Straightforward practice. Get the method working first.
5 problemsIntegral of 3x² from 1 to 3, antiderivative x³. Value?
Answer 26
Why 26.
Why does a definite integral need no + C?
Answer It appears in both terms and cancels in the subtraction.
Why The constants cancel.
Integral of 2x from 0 to 4, antiderivative x². Value?
Answer 16
Why 16 − 0.
Integral of 3x² from 0 to 2, antiderivative x³. Value?
Answer 8
Why 8 − 0.
Integral of 2x from 1 to 3, antiderivative x². Value?
Answer 8
Why 9 − 1.
Build It Up
The same ideas with more to keep track of.
3 problemsIntegral of 4 from 0 to 5, antiderivative 4x. Value?
Answer 20
Why 20 − 0.
An integral from 2 to 7 is 12. What is it from 7 to 2?
Answer -12
Why Swapping the limits negates it.
Integral from 0 to 3 is 5, and from 3 to 8 is 4. What is it from 0 to 8?
Answer 9
Why Areas add end to end.
Stretch Yourself
Mixed problems. Work out what kind of question it is before you start.
4 problemsWhich statements about definite integrals are true?
Answer Area below the axis counts as negative; Swapping the limits negates the value
Why The constant cancels in the subtraction.
Integral of 2x from 0 to 5, antiderivative x². Value?
Answer 25
Why 25 − 0.
The Evaluation: Integral of 2x from 0 to 3. An antiderivative is x². What is the value?
Answer 9
Why 9.
The Same Point: Integral of any function from 5 to 5. What is the value?
Answer 0
Why 0.