The Two Fundamental Theorems, Kept Straight

You’ll be able to

State both parts of the Fundamental Theorem of Calculus and their distinct uses
Differentiate an accumulation function, including with a variable inside the upper limit
Recognize which part a question requires

Two different statements with the same name

Part 1, the evaluation part, is the computational workhorse: ∫ₐᵇ f(x)dx = F(b) − F(a) where F is any antiderivative of f. Part 2, the derivative part, says that accumulation and differentiation undo each other: if g(x) = ∫ₐˣ f(t)dt then g′(x) = f(x). They answer opposite questions. Part 1 turns an integral into arithmetic on an antiderivative; Part 2 turns a derivative of an integral back into the integrand without ever computing the integral. Students who blur them attempt to evaluate an integral they were asked to differentiate — a great deal of work leading to the wrong kind of answer.

Part 2 with a chain rule

When the upper limit is a function of x rather than x itself, Part 2 comes with a chain-rule factor: if g(x) = ∫ₐ^(u(x)) f(t)dt then g′(x) = f(u(x))·u′(x). So d/dx ∫₀^(x²) sin(t)dt = sin(x²)·2x. If the lower limit carries the variable, the sign flips, because swapping limits negates an integral: d/dx ∫ₓ^b f(t)dt = −f(x). If both limits are functions, apply the rule to each and subtract. The constant limit contributes nothing, which is the reason a specific number appears in one limit at all — it fixes the value g(a) = 0 and is otherwise inert.

Accumulation functions have shape too

Given g(x) = ∫ₐˣ f(t)dt, everything Unit 5 says about shape applies with g′ = f. So g is increasing where f is positive, decreasing where f is negative, and has a local extremum where f changes sign. Since g″ = f′, g is concave up where f is increasing and has an inflection point where f changes from increasing to decreasing. This is why the exam so often gives a graph of f and asks about g: it is testing whether you can shift every question up one derivative. And g(a) = 0 always, since the integral from a to a is zero — which is frequently the only value of g you are given and is enough to pin the rest down.

FTC Part 2 with chain rule

d/dx ∫ₐ^(u(x)) f(t)dt = f(u(x)) · u′(x)

Substitute the upper limit into the integrand, then multiply by its derivative. A variable lower limit contributes the same thing with a minus sign.

Worked example

Let g(x) = ∫₁^(x³) √t dt. Find g′(x), and find g′(2).

1.The integrand is f(t) = √t and the upper limit is u(x) = x³.
2.By FTC Part 2 with the chain rule: g′(x) = f(u(x))·u′(x) = √(x³)·3x².
3.Simplify: √(x³) = x^(3/2) for x ≥ 0, so g′(x) = 3x²·x^(3/2) = 3x^(7/2).
4.Evaluate at x = 2: g′(2) = 3·2^(7/2).
5.2^(7/2) = 2³·√2 = 8√2 ≈ 11.3137, so g′(2) = 3(8√2) = 24√2 ≈ 33.941.

Answer: g′(x) = 3x^(7/2), and g′(2) = 24√2 ≈ 33.941. Note that no antiderivative of √t was ever computed — Part 2 makes that unnecessary.

On the exam

If a question shows an integral and asks for a derivative, reach for Part 2 and do not integrate. Students who evaluate the integral first can still reach the right answer on simple integrands, but they lose the time and they fail entirely when the integrand has no elementary antiderivative — which the exam includes deliberately.

Checkpoint

If g(x) = ∫₂ˣ f(t)dt and f is continuous, then g′(x) equals

Checkpoint

g(x) = ∫₀ˣ f(t)dt, where the graph of f is positive on (0, 4) and negative on (4, 7). At x = 4, g has

Answer the 2 checkpoints as you read.