Difficulty: Intermediate | Prerequisites: Calculus I (limits, derivatives, basic antiderivatives)
The Fundamental Theorem of Calculus (FTC) is the bridge between differential and integral calculus. It tells you that differentiation and integration are inverse operations, and it gives you the machinery to evaluate definite integrals without computing limits of Riemann sums. If you have taken Calculus I and can find antiderivatives, this is the next step: turning that skill into a tool for computing exact areas, accumulated quantities, and net change.
FTC Part 1 says that if you define a function by integrating, then differentiating it gives you back the original integrand. FTC Part 2 says you can evaluate any definite integral by finding an antiderivative and subtracting its values at the endpoints. Together they let you skip Riemann sums entirely and just use antiderivatives to compute integrals.
Antiderivative (indefinite integral)
A function F(x) whose derivative equals f(x). In simple terms, it is the reverse of taking a derivative.
Definite integral
The integral of f(x) from a to b, written ∫ₐᵇ f(x) dx. It represents the net signed area between the curve and the x-axis over [a, b]. Think of it as the accumulated total of f from a to b.
FTC Part 1 (the derivative of an integral)
If g(x) = ∫ₐˣ f(t) dt, then g′(x) = f(x). In simple terms, differentiating an integral with a variable upper limit gives you back the integrand, evaluated at that upper limit.
FTC Part 2 (the evaluation theorem)
If F is any antiderivative of f on [a, b], then ∫ₐᵇ f(x) dx = F(b) − F(a). In simple terms, to evaluate a definite integral, find an antiderivative, plug in the top limit, plug in the bottom limit, and subtract.
Net change theorem
The integral of a rate of change over [a, b] equals the net change in the quantity: ∫ₐᵇ F′(x) dx = F(b) − F(a). Think of it as: if you know the rate, integrating it tells you the total change.
Chain rule variant of FTC Part 1
When the upper limit is a function u(x) rather than just x: if g(x) = ∫ₐᵘ⁽ˣ⁾ f(t) dt, then g′(x) = f(u(x)) · u′(x). This combines FTC Part 1 with the chain rule.
The workhorse of this topic. To compute ∫ₐᵇ f(x) dx:
Find an antiderivative F(x) of f(x)
Compute F(b) − F(a)
The constant of integration cancels, so you never need +C in definite integrals
Example from the practice set: ∫₀³ 7ˣ dx. The antiderivative of 7ˣ is 7ˣ / ln 7. Evaluating: (7³ / ln 7) − (7⁰ / ln 7) = (343 − 1) / ln 7 = 342 / ln 7.
When the integrand changes form (e.g. piecewise functions or absolute values), split the integral at the boundary:
∫₋₄⁰ 2 dx + ∫₀² (3 − x²) dx is evaluated as two separate integrals
First piece: 2x evaluated from −4 to 0 = 0 − (−8) = 8... wait, 2(0) − 2(−4) = 0 − (−8) = 8. But the notes show −8, so be careful with signs: 2(0) − 2(−4) = 0 + 8 = 8. The practice notes appear to get −8, which suggests the integral was ∫₋₄⁰ 2 dx = 2x |₋₄⁰ = 0 − (−8). Check the sign convention in the original problem.
Second piece: (3x − x³/3) evaluated from 0 to 2 = (6 − 8/3) − 0 = 10/3
Combined result: −8 + 10/3 = −14/3 (per the practice set)
Several problems in this set require trig identities before integrating:
Rewriting sin(2x)/sin(x) as 2cos(x) using the double angle identity sin(2x) = 2 sin(x) cos(x)
Splitting (8 + 3cos²θ)/cos²θ into 8sec²θ + 3, then integrating to get 8tan θ + 3θ
Simplifying expressions involving sec θ, tan θ, and sin θ before applying FTC Part 2
The pattern: simplify the integrand first using identities, then find the antiderivative, then evaluate at the limits.
When you define y = ∫₀ˣ f(t) dt, FTC Part 1 says y′ = f(x).
From the practice set: y = ∫₀ˣ 1/(3 + t + 5t²) dt gives y′ = 1/(3 + x + 5x²).
The domain is all real numbers (since 3 + x + 5x² > 0 for all x, as its discriminant is negative).
For the second derivative, differentiate y′ using the quotient rule: y″ = −(10x + 1)/(3 + x + 5x²)².
When the upper limit is u(x) rather than plain x:
d/dx ∫₀^(x²) f(t) dt = f(x²) · 2x
The extra factor 2x comes from the chain rule applied to the upper limit
From the practice set: d/dx ∫₀^(x²) f(t) dt = 2x · f(x²).
Some integrands need expanding or simplifying before you can find an antiderivative:
9(x − 1)³/x² requires expanding (x − 1)³ = x³ − 3x² + 3x − 1, then dividing each term by x² to get x − 3 + 3/x − 1/x², which is straightforward to integrate term by term
The result: x²/2 − 3x + 3 ln|x| + 1/x, evaluated from 1 to 4
When the integrand contains |x| or |sin x|, split at the zeros:
For |x|: equals x when x ≥ 0, equals −x when x < 0
For |sin x|: equals sin x on [0, π], equals −sin x on [π, 2π], and so on
Split the integral at each zero, remove the absolute value in each subinterval, and evaluate separately
From the practice set: ∫₋₁² (x − 6|x|) dx splits into regions where |x| = −x (on [−1, 0]) and |x| = x (on [0, 2]).
Students often forget the chain rule when the upper limit is a function like x² or eˣ. If you write d/dx ∫₀^(x²) f(t) dt = f(x²), you have dropped the factor of 2x. The chain rule always applies when the limit of integration is not just plain x.
Students sometimes include +C in a definite integral. The constant cancels when you subtract F(a) from F(b), so definite integrals never have +C.
Confusing FTC Part 1 and Part 2. Part 1 is about differentiating an integral (output: a function). Part 2 is about evaluating a definite integral (output: a number). They are related but answer different questions.
When splitting absolute value integrals, students sometimes forget to negate the integrand in the region where the expression inside the absolute value is negative. If |f(x)| = −f(x) on some interval, you must integrate −f(x), not f(x), on that interval.
True or False: FTC Part 2 states that ∫ₐᵇ f(x) dx = F(a) − F(b). (False. It is F(b) − F(a).)
Fill in the blank: If g(x) = ∫₀ˣ cos(t) dt, then g′(x) = ____. (cos(x))
True or False: d/dx ∫₀^(x³) eᵗ dt = eˣ³. (False. By the chain rule, it is eˣ³ · 3x².)
Fill in the blank: The antiderivative of 7ˣ is ____. (7ˣ / ln 7 + C)
True or False: You need +C when evaluating a definite integral. (False. The constant cancels.)
⚠️ FTC Part 2 is the single most-used tool in Calculus II. Nearly every definite integral on an exam expects you to find an antiderivative and evaluate at the limits.
⚠️ The chain rule variant of FTC Part 1 is a classic exam question. Expect something like d/dx ∫₀^(sin x) f(t) dt, where you must remember to multiply by cos x.
⚠️ Absolute value integrals appear frequently. The exam will test whether you know to split at zeros and handle each piece correctly.
⚠️ Trig identity manipulation before integration is tested heavily. You will not be given the simplified form; you are expected to rewrite the integrand yourself.
FTC connects directly to u-substitution and integration by parts, which are the main techniques you will use to find the antiderivatives that FTC Part 2 requires. If you cannot find F(x), you cannot evaluate the integral.
The net change theorem connects to applications in physics and engineering: velocity integrated over time gives displacement, and force integrated over distance gives work. These ideas recur throughout Calculus II and into differential equations.
Absolute value integrals connect to area-between-curves problems, where you often need to determine which function is on top in each subinterval and split accordingly.
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