Fundamental Theorem of Calculus
In short: Fundamental theorem of calculus links the two big operations: differentiating an accumulated area returns the original function, and a definite integral equals the antiderivative at the top minus at the bottom.
AP Calculus ABCalculusAP Calculus AB Unit 6
What is Fundamental Theorem of Calculus?
The fundamental theorem of calculus is the bridge between slopes and areas. It says that accumulating a quantity and then differentiating brings you back to where you started.
Part 1: if \(f\) is continuous and \(A(x) = \int_a^x f(t)\,dt\), then \(A^{\prime}(x) = f(x)\). Part 2: if \(F\) is any antiderivative of \(f\), then \(\int_a^b f(x)\,dx = F(b) – F(a)\). With the second part, areas no longer require adding thousands of rectangles.
How to spot it
- A function is defined as an integral with a variable upper limit, like g(x) equals the integral from 2 to x of something.
- You are asked for g'(x) of such a function without evaluating the integral.
- The question wants the exact value of a definite integral using an antiderivative.
- The upper limit is an expression like x squared, which signals the chain rule is also needed.
The method
- For Part 2, find an antiderivative \(F\) and compute \(F(b) – F(a)\).
- For Part 1, replace the variable in the integrand by the upper limit: \(\frac{d}{dx}\int_a^x f(t)\,dt = f(x)\).
- If the upper limit is a function \(u(x)\), multiply by its derivative: \(f(u(x))\cdot u^{\prime}(x)\).
- Check that the integrand is continuous on the interval before applying the fundamental theorem of calculus.
Fundamental Theorem of Calculus: solved example
Problem
Let \(g(x) = \int_2^{x^3} (t^2 + 1)\,dt\). Find \(g^{\prime}(1)\), then evaluate \(\int_1^3 (4x – 1)\,dx\).
Solution
Part 1 with the chain rule: \(g^{\prime}(x) = \left((x^3)^2 + 1\right)\cdot 3x^2 = 3x^2(x^6 + 1)\). At \(x = 1\): \(3 \cdot 2 = \) 6.
Part 2: an antiderivative of \(4x – 1\) is \(2x^2 – x\). Then \((18 – 3) – (2 – 1) = 15 – 1 = \) 14.
Try it
Question 1
If \(h(x) = \int_0^x \sqrt{t^2 + 9}\,dt\), find \(h^{\prime}(4)\).
Show the answer
\(h^{\prime}(x) = \sqrt{x^2 + 9}\), so \(h^{\prime}(4) = \sqrt{25} = 5\).
Question 2
Evaluate \(\int_1^8 x^{1/3}\,dx\).
Show the answer
\(\left[\frac{3}{4}x^{4/3}\right]_1^8 = \frac{3}{4}(16) – \frac{3}{4}(1) = 12 – 0.75 = 11.25\).
Fundamental Theorem of Calculus vs. Definite Integral
A definite integral is a number you compute. The fundamental theorem is the statement that explains how to compute it, using antiderivatives, and also how to differentiate a function defined by an integral. One is the object; the other is the rule that connects it to derivatives.
Keep going
Lessons
Worksheets and quizzes
- Fundamental Theorem of Calculus and Antiderivatives – Worksheet
- Fundamental Theorem of Calculus and Antiderivatives – Quiz
Related words
Whole dictionary: A–Z · AP Calculus AB · Calculus