To evaluate an integral from a to b, find an antiderivative F, then compute F(b) − F(a).
Step 1: Let's Learn
Read it, or press Listen and follow the words.
The constant cancels
Any antiderivative works, since a + C on both terms subtracts away. That is why definite integrals never carry one.
Signed area
Area below the axis counts as negative. An integral can therefore be zero even when the region has real size.
Total area is different
For total area regardless of sign, split at every crossing and add the absolute values.
Swapping the limits
Reversing the limits negates the integral, since F(a) − F(b) is the negative of F(b) − F(a).
Splitting the interval
An integral from a to c equals the one from a to b plus the one from b to c. Areas add end to end.
Antiderivative at the top, minus at the bottom
Find any antiderivative F, then compute F(b) − F(a). The constant of integration cancels in the subtraction, which is why definite integrals need no C while indefinite ones do.
Substitution and the limits
When substituting u = g(x), either convert the limits to u values or convert back to x before evaluating. Substituting new limits into the old variable is the standard error and gives a confidently wrong number.
Useful properties
Reversing the limits negates the integral; splitting the interval splits the integral; a constant factor comes out. These let you break a hard integral into manageable pieces before any antiderivative is found.
Check by differentiating
Differentiate your antiderivative and confirm it returns the integrand. This check is complete and fast, and it is the natural one given that integration is defined as the reverse of differentiation.
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 Rosa Evaluate an Integral
Rosa integrates 3x² from 1 to 2.
- Step 1
An antiderivative of 3x² is x³.
Step 4: Your Turn
Practice makes it stick.
The Evaluation
Problem 1 of 2
Integral of 2x from 0 to 3. An antiderivative is x². What is the value?
The Same Point
Problem 2 of 2
Integral of any function from 5 to 5. What is the value?
Top Minus Bottom
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
Integral of 2x from 1 to 3, antiderivative x². Value?
4 of 8
Integral of 4 from 0 to 5, antiderivative 4x. Value?
5 of 8
An integral from 2 to 7 is 12. What is it from 7 to 2?
6 of 8
Integral from 0 to 3 is 5, and from 3 to 8 is 4. What is it from 0 to 8?
7 of 8
Which statements about definite integrals are true?
8 of 8
Integral of 2x from 0 to 5, antiderivative x². Value?
Step 5: Quick Check
Show what you know.
Question 1 of 2
Integral of 3x² from 1 to 3, antiderivative x³. Value?
Question 2 of 2
Why does a definite integral need no + C?
What You Learned
- 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.