The Fundamental Theorem of Calculus
Differentiation and integration look like unrelated ideas. This theorem shows that one undoes the other. Drag the sliders below to see why before you see the statement.
Area as a function
Fix a starting point at t = 0 and let A(x) be the signed area under f from 0 to x. As x moves right, the area grows, so A is itself a function. How fast does it grow?
The orange sliver is the extra area gained between x and x + h. Shrink h and watch the two numbers converge. On the lower graph, the tangent’s slope at the dot is always f(x).
The theorem
Fundamental Theorem of Calculus, Part 1. If f is continuous on [a, b] and
g(x) = ∫ax f(t) dt
then g′(x) = f(x) for every x in (a, b).
In words: the rate at which accumulated area grows equals the height of the curve at the edge. That is exactly what the sliver showed. For a tiny h, the sliver is nearly a rectangle of height f(x) and width h, so its area divided by h is about f(x).
Three rules that come from it
1. Plain upper limit. d/dx ∫ax f(t) dt = f(x). Substitute x into the integrand.
2. Variable lower limit. Swapping limits flips the sign, so d/dx ∫xb f(t) dt = −f(x).
3. Chain rule. d/dx ∫au(x) f(t) dt = f(u(x)) · u′(x).
Worked example
Let g(x) = ∫0x √(t² + 5) dt. Find g′(2).
You don’t need to evaluate the integral, which has no simple antiderivative. By the theorem, g′(x) = √(x² + 5), so g′(2) = √9 = 3.
Your turn
Score: 0 of 6 correct
Part 1 says the derivative of an area function is the original function. Part 2 turns that around into a shortcut: to find an exact area, find an antiderivative and subtract.
Fundamental Theorem of Calculus, Part 2. If f is continuous on [a, b] and F′ = f (so F is any antiderivative of f), then
∫ab f(x) dx = F(b) − F(a)
Evaluate the antiderivative at the top, subtract its value at the bottom.
Why it follows from Part 1
Part 1 tells us that A(x) = ∫ax f(t) dt is one antiderivative of f. Any other antiderivative F differs from it by a constant, F = A + C. Then F(b) − F(a) = (A(b) + C) − (A(a) + C) = A(b) − A(a), and since A(a) = 0 this is exactly the area from a to b. The constant cancels.
Test it
Drag a and b to choose an interval. The first number measures the shaded area numerically. The second comes from the antiderivative. Then slide C and notice the answer doesn’t move.
Worked example
Evaluate ∫13 (2x + 1) dx.
An antiderivative is F(x) = x² + x. Then F(3) − F(1) = 12 − 2 = 10.
Your turn
Score: 0 of 4 correct
Most real uses of integrals start with a rate: gallons per minute, dollars per unit, pounds per year. This tab is about reading the integral of a rate in plain language.
Net change. If f′ is the rate of change of a quantity f, then
∫ab f′(t) dt = f(b) − f(a)
Adding up a rate over an interval gives the total change in the quantity over that interval. This is Part 2 with F = f.
Explaining an integral in words
Ask three questions. What quantity does the integrand measure the rate of? Over what interval? And what are the units? The units of the integral are the units of the rate times the units of t, which are the units of the quantity itself.
Worked example
Let f′(t) be the rate of growth of a child in pounds per year, where t is age in years. Explain the meaning of ∫48 f′(t) dt.
Answer. The integral adds up the child’s growth over every moment from age 4 to age 8. By the theorem it equals f(8) − f(4), so it is the total change in the child’s weight, in pounds, between ages 4 and 8: how much weight the child gains from the 4th to the 8th birthday.
It is not the child’s weight at age 8, which would also need the weight at age 4. It is not a rate either: its units are (pounds/year) × (years) = pounds. Dividing by 4 years would give the average growth rate.
Your turn
Score: 0 of 6 correct
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