Key takeaway: The fundamental theorem of calculus links derivatives and integrals: the derivative of an accumulation function returns the integrand, and a definite integral equals an antiderivative evaluated at the upper limit minus the lower limit.
Grade 12 – AP Calculus ABAP Calculus AB Unit 6Calculus
Definition of Fundamental Theorem of Calculus
The fundamental theorem of calculus (FTC) connects the two big ideas of the course, slopes and areas. It shows they are inverse processes: building up area and then taking the slope of that area brings back the original function.
It has two parts. One says how to differentiate a function defined by an integral with a variable upper limit. The other turns the hard limit of Riemann sums into subtraction using any antiderivative.
Formula and notation
Let \(r\) be continuous and \(R\) any antiderivative of \(r\).
\[ \frac{d}{dx}\int_{m}^{x} r(t)\,dt=r(x),\qquad \int_{m}^{n} r(x)\,dx=R(n)-R(m). \]
The first formula is Part 1 and the second is Part 2.
Key facts
- In Part 1 the variable upper limit becomes the input of the integrand.
- The lower limit in an accumulation function is a constant and does not appear in the derivative.
- The fundamental theorem of calculus, Part 2, lets you skip Riemann sums and use any antiderivative.
- With a chain-rule factor, an upper limit like x squared needs the derivative of that limit as well.
Fundamental Theorem of Calculus example
Let \(g(x)=\int_{3}^{x}(t^2+4t)dt\). Find \(g^{\prime}(5)\) and \(g(3)\).
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By Part 1, the derivative of the accumulation function is the integrand evaluated at \(x\): \(g^{\prime}(x)=x^2+4x\).
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At \(x=5\): \(g^{\prime}(5)=25+20=45\).
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An integral from 3 to 3 covers no width, so \(g(3)=0\).
The derivative at 5 is \(g^{\prime}(5)=45\), and the starting value is \(g(3)=0\).
Common mistake
Watch out
A student subtracts the lower limit’s output, writing \(g^{\prime}(x)=(x^2+4x)-(3^2+4\cdot 3)\) and getting \(45-21=24\) at \(x=5\).
Why it is wrong
The lower limit of an accumulation function is a fixed number, and it only shifts the function \(g\) by a constant, which vanishes in the derivative. Part 1 of the fundamental theorem of calculus yields just the integrand at \(x\): \(g^{\prime}(5)=45\).
Related terms and lessons
Browse the glossary: A–Z list · AP Calculus AB vocabulary · Calculus