How can an antiderivative evaluate a definite integral?
Quick answer
Find an antiderivative F of the integrand f, then calculate F(b) − F(a), where a is the lower endpoint and b is the upper endpoint. For example, ∫₁³(2x + 1) dx = [x² + x]₁³ = (9 + 3) − (1 + 1) = 10. Keep the endpoint order: upper value first, lower value second.
The question, the details, the next step.
I know how to find an antiderivative, but how does it give me the value of a definite integral?
For a continuous function f on an interval [a,b], the evaluation rule is ∫ₐᵇ f(x) dx = F(b) − F(a), where F is an antiderivative of f. In plain language: find a function whose derivative is the integrand, evaluate that function at the upper endpoint, then subtract its value at the lower endpoint. [S1]
Try this example: evaluate ∫₁³(2x + 1) dx.
- Find an antiderivative: since the derivative of x² + x is 2x + 1, take F(x) = x² + x.
- Apply the bounds: [x² + x]₁³ means F(3) − F(1).
- Substitute and simplify: (3² + 3) − (1² + 1) = 12 − 2 = 10.
So the definite integral is 10. The interval is from 1 to 3; those bounds tell you which two values of the antiderivative to compare.
Why do I subtract the lower endpoint from the upper one, instead of the other way around?
The rule specifically uses F(b) − F(a): the value at the upper endpoint minus the value at the lower endpoint. [S1] In this example, the upper endpoint is 3 and the lower endpoint is 1, so the calculation is F(3) − F(1), or 12 − 2.
A useful way to avoid reversing the order is to write the endpoint values before doing arithmetic: “upper: F(3) = 12; lower: F(1) = 2.” Then write upper minus lower. If you swap the order, you get the opposite sign, not the requested integral with its bounds in the original order.
What does the bracket notation [x² + x]₁³ mean? Do I use the antiderivative’s constant too?
The notation [x² + x]₁³ is shorthand for “substitute 3, then subtract the result of substituting 1”: (3² + 3) − (1² + 1). It is not asking you to integrate again; it records the two endpoint evaluations and their order.
You can write a constant in an antiderivative, such as F(x) = x² + x + C. It cancels when you subtract: (9 + 3 + C) − (1 + 1 + C) = 12 − 2 = 10. So it is usually convenient to leave the constant out when evaluating a definite integral.
Can you give me a reusable setup so I can check my own answer?
Use this template, replacing the letters and expressions with your problem:
∫ₐᵇ f(x) dx = [F(x)]ₐᵇ = F(b) − F(a) = (value at upper endpoint) − (value at lower endpoint)
For the worked example, that becomes ∫₁³(2x + 1) dx = [x² + x]₁³ = (3² + 3) − (1² + 1) = 10.
Before you finish, check that F really differentiates to the integrand, that you substituted both stated bounds, and that the upper-endpoint value comes first. The rule cited here applies when the integrand is continuous across the interval. [S1]
A practical checklist
Identify the integrand and the lower and upper endpoints.
Find an antiderivative F and check that its derivative matches the integrand.
Write F(upper endpoint) − F(lower endpoint), in that order.
Substitute each endpoint into the whole antiderivative and simplify.
Check the interval and the sign of your final result.
A few more useful answers
What if the upper and lower endpoints are reversed?
Keep the bounds as written when applying the rule: evaluate at the upper endpoint and subtract the value at the lower endpoint. Reversing the bounds reverses the subtraction order.
Does the constant of integration change the answer?
No. The same constant appears in both endpoint evaluations and cancels when you subtract, so you can omit it for this calculation.
What does the definite integral’s answer represent?
This method evaluates the signed definite integral using endpoint values of an antiderivative. The evaluation rule applies when the integrand is continuous across the interval. [S1]
Sources
References used for this guide.
- [S1] The Fundamental Theorem of Calculus, Part 2 — OpenStax Calculus Volume 2 openstax.org · Checked