Difficulty: Intermediate | Prerequisites: Basic integration concepts, chain rule for derivatives, trigonometric functions
The Fundamental Theorem of Calculus Part I is the bridge between integration and differentiation. It tells you how to differentiate a function defined as a definite integral with a variable upper limit. This topic sits at the heart of Calculus I and shows up heavily on exams, particularly in AP Calculus AB/BC free-response sections. You should already be comfortable with the chain rule and basic antiderivative notation before tackling this.
If g(x) is defined as a definite integral with x in the limits, FTC Part I lets you find g'(x) without ever computing the integral itself. You substitute x (or whatever function of x sits in the limit) into the integrand, then multiply by the derivative of that limit. When the variable is in the lower limit, you pick up a negative sign.
Fundamental Theorem of Calculus Part I (FTC I)
If f is continuous on [a, b] and g(x) = ∫ from a to x of f(t) dt, then g'(x) = f(x). In simple terms, differentiating an integral "undoes" the integration: you get the integrand back, evaluated at the upper limit.
Accumulation function
A function defined as a definite integral with a variable limit, such as g(x) = ∫ from a to x of f(t) dt. Think of it as a running total: g(x) measures the net area under f from a up to x.
Integrand
The function being integrated, the f(t) inside the integral sign. When applying FTC I, you substitute the upper limit into the integrand.
Dummy variable
The variable of integration (usually t) that disappears after integration. It is not the same as x. In simple terms, t is a placeholder; the answer depends on x, not t.
Chain rule with FTC I
When the upper limit is not simply x but some function u(x), the derivative picks up an extra factor of u'(x). Think of it as: differentiate the outer integral (FTC I gives you f(u)), then multiply by the derivative of the inner function (u'(x)).
When g(x) = ∫ from a to x of f(t) dt, apply FTC I directly: g'(x) = f(x).
g(x) = ∫ from 0 to x of (1 + t²) dt → g'(x) = 1 + x²
g(x) = ∫ from π to x of (2 + cos t) dt → g'(x) = 2 + cos x
All you do is replace every t in the integrand with x.
When the upper limit is a function of x rather than x itself, the chain rule applies: g'(x) = f(u(x)) · u'(x).
g(x) = ∫ from 2 to 1/x of sin⁴(t) dt, where u = 1/x and u' = −1/x²
g'(x) = sin⁴(1/x) · (−1/x²)
g(x) = ∫ from 1 to √x of (t²/(t²+1)) dt, where u = √x and u' = 1/(2√x)
g'(x) = (x/(x+1)) · 1/(2√x)
The critical step: identify u(x) in the upper limit, find u'(x), then multiply.
When x appears in the lower limit rather than the upper, flip the integral first:
∫ from u(x) to a of f(t) dt = –∫ from a to u(x) of f(t) dt
Then apply the chain rule as normal. The flip introduces a negative sign.
g(x) = ∫ from tan(x) to 17 of sin(t⁴) dt = –∫ from 17 to tan(x) of sin(t⁴) dt
g'(x) = –sin(tan⁴(x)) · sec²(x)
When both limits depend on x, split the integral at any constant c:
∫ from u(x) to v(x) of f(t) dt = ∫ from u(x) to c of f(t) dt + ∫ from c to v(x) of f(t) dt
Differentiate each piece separately, applying the chain rule to each.
d/dx ∫ from cos(x) to 2 of 4/(1+t³) dt: the upper limit is constant (2) and the lower limit is cos(x)
Flip: –∫ from 2 to cos(x) of 4/(1+t³) dt
g'(x) = –[4/(1+cos³(x))] · (–sin x) = 4 sin(x)/(1+cos³(x))
d/dx ∫ from 9 to e²ˣ of √(1+t³) dt: upper limit is e²ˣ, lower is constant (9)
g'(x) = √(1+e⁶ˣ) · 2e²ˣ
Basic FTC I
If g(x) = ∫ from a to x of f(t) dt, then g'(x) = f(x).
FTC I with Chain Rule
If g(x) = ∫ from a to u(x) of f(t) dt, then g'(x) = f(u(x)) · u'(x).
Variable Lower Limit
If g(x) = ∫ from u(x) to a of f(t) dt, then g'(x) = –f(u(x)) · u'(x).
Both Limits Variable
If g(x) = ∫ from u(x) to v(x) of f(t) dt, then g'(x) = f(v(x)) · v'(x) – f(u(x)) · u'(x).
Students often try to evaluate the integral first and then differentiate. You do not need to find an antiderivative. FTC I skips that step entirely.
Students forget the chain rule factor when the upper limit is something other than plain x. If the upper limit is √x, you must multiply by d/dx(√x) = 1/(2√x).
Students drop the negative sign when the variable is in the lower limit. Flipping the limits of integration always introduces a factor of −1.
Students confuse the dummy variable t with the variable x. After applying FTC I, no t should remain in your answer. Every t is replaced by whatever sits in the variable limit.
⚠️ FTC I with composite upper limits (chain rule) is one of the most commonly tested topics on AP Calculus exams. Expect at least one free-response part requiring it.
⚠️ Problems that combine FTC I with concavity analysis (finding where g is concave up/down) appear regularly. These require taking the second derivative, meaning you differentiate the integrand itself.
⚠️ Trig functions as upper or lower limits (sin x, tan x, cos x) are a favourite exam device because they test both FTC I and trig derivative recall simultaneously.
⚠️ Variable in both limits is less common but always appears as a harder problem. Splitting the integral at a constant is the standard technique.
True or false: To use FTC I, you must first find an antiderivative of the integrand. (False)
Fill in the blank: If g(x) = ∫ from 0 to x of cos(t²) dt, then g'(x) = ____. (cos(x²))
True or false: If the variable x is in the lower limit, you need a negative sign. (True)
Fill in the blank: If g(x) = ∫ from 1 to x³ of e^t dt, then g'(x) = e^(x³) · ____. (3x²)
True or false: After applying FTC I, the dummy variable t may still appear in the final answer. (False)
Q: Find g'(x) if g(x) = ∫ from 0 to x of (1 + t²) dt.
A: g'(x) = 1 + x². Direct FTC I: substitute x into the integrand.
Q: Find g'(x) if g(x) = ∫ from 1 to √x of (t²/(t² + 1)) dt.
A: g'(x) = (x/(x + 1)) · 1/(2√x). The upper limit is √x, so apply FTC I to get (√x)²/((√x)² + 1) = x/(x + 1), then multiply by the derivative of √x, which is 1/(2√x).
Q: Evaluate d/dx ∫ from cos(x) to 2 of 4/(1 + t³) dt.
A: 4 sin(x)/(1 + cos³(x)). Flip the integral to get –∫ from 2 to cos(x) of 4/(1 + t³) dt. The derivative is –[4/(1 + cos³(x))] · (–sin x) = 4 sin(x)/(1 + cos³(x)).
Q: Find g'(x) if g(x) = ∫ from −5 to sin(x) of t cos(t³) dt.
A: g'(x) = sin(x) · cos(sin³(x)) · cos(x). Substitute sin(x) for t in the integrand to get sin(x) · cos(sin³(x)), then multiply by the derivative of sin(x), which is cos(x).
Q: Evaluate d/dx ∫ from 9 to e²ˣ of √(1 + t³) dt.
A: √(1 + e⁶ˣ) · 2e²ˣ. Substitute e²ˣ for t: √(1 + (e²ˣ)³) = √(1 + e⁶ˣ). Multiply by the derivative of e²ˣ, which is 2e²ˣ.
Fundamental Theorem of Calculus, FTC Part 1, FTC I, differentiation under the integral sign, Leibniz integral rule, accumulation function, integral with variable limits, chain rule with integrals, derivative of an integral, AP Calculus AB free response, AP Calculus BC, MSU Calculus I, Michigan State MTH 132, variable upper limit, variable lower limit, composite function chain rule, integrand substitution