Course: MTH 132, Calculus I (Michigan State University) Source: Recitation Activity 13, Sections 4.3, 4.4 & 5.5 Difficulty: Intermediate Prerequisites: Basic derivative rules (power, product, chain), antiderivative formulas, understanding of summation notation and Riemann sums.
The Fundamental Theorem of Calculus (FTC) connects differentiation and integration: Part 1 tells you how to differentiate a function defined as an integral, and Part 2 lets you evaluate definite integrals using antiderivatives. Most of this recitation is about applying these two ideas, plus the chain rule when the limits of integration are functions of x.
You have spent weeks learning how to take derivatives and how to find antiderivatives. The Fundamental Theorem of Calculus is the bridge between those two worlds: it says that differentiation and integration are inverse operations. Section 4.3 (FTC Part 1) shows you how to differentiate an integral with a variable limit. Section 4.4 (FTC Part 2, also called the Evaluation Theorem) gives you the tool to compute exact values of definite integrals without limits of Riemann sums. Section 5.5 ties these to motion problems (distance, displacement) and average value.
If you are comfortable with basic antiderivative rules and the chain rule, you are ready for this material. If not, review those first.
Fundamental Theorem of Calculus, Part 1 (FTC1)
If f is continuous on [a, b], then the function F(x) = ∫ from a to x of f(t) dt is differentiable, and F'(x) = f(x). In simple terms, differentiating an integral with a variable upper limit just gives you back the integrand evaluated at that limit.
Fundamental Theorem of Calculus, Part 2 (FTC2 / Evaluation Theorem)
If f is continuous on [a, b] and F is any antiderivative of f, then ∫ from a to b of f(x) dx = F(b) − F(a). Think of it as: to evaluate a definite integral, find any antiderivative, plug in the top limit, plug in the bottom limit, and subtract.
Definite integral
The integral ∫ from a to b of f(x) dx, which produces a number (not a function). It represents the net signed area between f(x) and the x-axis on [a, b].
Antiderivative
A function F(x) whose derivative is f(x). Also written as the indefinite integral ∫ f(x) dx = F(x) + C.
Net signed area
Area above the x-axis counts as positive, area below counts as negative. The definite integral gives the net of these. In simple terms, if the curve dips below the axis, that region subtracts from the total.
Average value of a function
For f continuous on [a, b], the average value is (1/(b − a)) · ∫ from a to b of f(x) dx. Think of it as the height of a rectangle with the same base [a, b] that has the same area as the region under f.
When F(x) = ∫ from a to x of f(t) dt, the derivative is simply F'(x) = f(x). You replace the dummy variable with the upper limit.
Example: f(y) = ∫ from 2 to y of t² sin(t) dt
By FTC1, f'(y) = y² sin(y)
You just substitute y for t in the integrand
If the upper limit is g(x) instead of just x, then you need the chain rule:
d/dx [∫ from a to g(x) of f(t) dt] = f(g(x)) · g'(x)
You evaluate the integrand at the upper limit, then multiply by the derivative of that upper limit.
Split the integral at any convenient constant (often 0), then differentiate each piece:
∫ from h(x) to g(x) of f(t) dt = ∫ from 0 to g(x) of f(t) dt − ∫ from 0 to h(x) of f(t) dt
Derivative = f(g(x)) · g'(x) − f(h(x)) · h'(x)
Example: g(x) = ∫ from 2x to 3x of 5/(u² + 5) du
Split: ∫ from 0 to 3x of 5/(u² + 5) du − ∫ from 0 to 2x of 5/(u² + 5) du
g'(x) = [5/((3x)² + 5)] · 3 − [5/((2x)² + 5)] · 2
g'(x) = 15/(9x² + 5) − 10/(4x² + 5)
Example: y(x) = ∫ from cos(x) to 5x of cos(v²) dv
Split: ∫ from 0 to 5x of cos(v²) dv − ∫ from 0 to cos(x) of cos(v²) dv
y'(x) = cos((5x)²) · 5 − cos((cos x)²) · (−sin x)
y'(x) = 5 cos(25x²) + sin(x) · cos(cos²x)
Note the sign flip: the derivative of cos(x) is −sin(x), and the subtraction introduces a second negative, making the second term positive
The process: find the antiderivative F(x), then compute F(b) − F(a). No constant of integration needed.
Example: ∫ from 2 to 4 of (6x² − 2x + 9) dx
Antiderivative: 2x³ − x² + 9x
Evaluate: [2(4)³ − (4)² + 9(4)] − [2(2)³ − (2)² + 9(2)]
= [128 − 16 + 36] − [16 − 4 + 18] = 148 − 30 = 118
Example: ∫ from 0 to 2 of (x + 2)(x² − 1) dx
First expand: x³ + 2x² − x − 2
Antiderivative: x⁴/4 + 2x³/3 − x²/2 − 2x
Evaluate at 2: 16/4 + 16/3 − 4/2 − 4 = 4 + 16/3 − 2 − 4 = −2 + 16/3 = 10/3
Evaluate at 0: 0
Result: 10/3
When the integrand involves fractions with powers of x, rewrite using negative and fractional exponents before integrating.
Example: ∫ from 1 to √7 of (2x + 6√x)/x⁴ dx
Rewrite: (2x)/x⁴ + (6x^(1/2))/x⁴ = 2x⁻³ + 6x⁻⁷ᐟ²
Antiderivative: −x⁻² − (12/5)x⁻⁵ᐟ²
Evaluate at the bounds and subtract to get approximately 3.05
Example: ∫ from 0 to π/4 of 5 sec²(x) dx
Antiderivative of sec²(x) is tan(x)
= 5 tan(x) evaluated from 0 to π/4
= 5(1) − 5(0) = 5
\frac{d}{dx}\left[\int_a^x f(t)\,dt\right] = f(x)\frac{d}{dx}\left[\int_a^{g(x)} f(t)\,dt\right] = f(g(x))\cdot g'(x)\frac{d}{dx}\left[\int_{h(x)}^{g(x)} f(t)\,dt\right] = f(g(x))\cdot g'(x) - f(h(x))\cdot h'(x)\int_a^b f(x)\,dx = F(b) - F(a) \quad\text{where } F'(x) = f(x)f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dxThe FTC is how engineers and scientists move between rates and totals. If you know a rate of change (flow rate, velocity, power), integrating it gives you the accumulated quantity (volume, displacement, energy). Conversely, if you have a cumulative measurement, differentiating recovers the instantaneous rate. This is the mathematical backbone of every "how much total" question in physics, economics and engineering.
Students often forget the chain rule when the upper limit is something other than plain x. If the upper limit is 3x, you must multiply by the derivative of 3x (which is 3). Missing this factor is the single most common error on FTC1 problems.
When both limits are functions of x, students sometimes try to differentiate the integrand directly instead of splitting the integral at a constant first. The split is essential.
Students sometimes add "+C" when evaluating a definite integral. The constant of integration cancels in the subtraction F(b) − F(a), so it is never written.
Confusing the dummy variable (t, u, v) with the variable of differentiation (x, y). The dummy variable disappears after you apply the FTC. It is just a placeholder.
⚠️ FTC1 with the chain rule (variable limits that are functions of x) is a near-certainty on any Calculus I exam covering this chapter. Expect at least one problem with a non-trivial upper limit.
⚠️ Evaluation of definite integrals using FTC2 is the backbone of the rest of the course. Every application problem (area, volume, work, average value) depends on this skill.
⚠️ Professors frequently test whether you can rewrite an integrand before integrating, particularly expressions involving roots and negative exponents.
⚠️ Know your trig antiderivatives cold: sec²(x) → tan(x), sin(x) → −cos(x), cos(x) → sin(x). These appear constantly.
True or False: d/dx [∫ from 0 to x² of sin(t) dt] = sin(x²).
True or False: ∫ from 1 to 3 of 2x dx = x² evaluated from 1 to 3 = 9 − 1 = 8.
Fill in the blank: To differentiate ∫ from cos(x) to x³ of e^t dt, you split the integral at ______ and apply the chain rule to each piece.
True or False: The "+C" is needed when evaluating a definite integral.
Fill in the blank: The antiderivative of sec²(x) is ______.
Q: Find the derivative of F(x) = ∫ from 1 to x³ of ln(t) dt.
A: By FTC1 with the chain rule, F'(x) = ln(x³) · 3x² = 3x² · 3 ln(x) = 9x² ln(x). Alternatively, ln(x³) = 3 ln(x), so F'(x) = 3 ln(x) · 3x² = 9x² ln(x).
Q: Evaluate ∫ from 0 to 2 of (3x² + 4x) dx.
A: Antiderivative is x³ + 2x². Evaluate: (8 + 8) − (0) = 16.
Q: Find the derivative of g(x) = ∫ from x to x² of e^(−t²) dt.
A: Split at 0: g(x) = ∫ from 0 to x² of e^(−t²) dt − ∫ from 0 to x of e^(−t²) dt. Then g'(x) = e^(−x⁴) · 2x − e^(−x²) · 1 = 2x·e^(−x⁴) − e^(−x²).
Q: Evaluate ∫ from 1 to 4 of (√x + 1/x²) dx.
A: Rewrite as x^(1/2) + x^(−2). Antiderivative: (2/3)x^(3/2) − x^(−1). At x = 4: (2/3)(8) − 1/4 = 16/3 − 1/4 = 61/12. At x = 1: 2/3 − 1 = −1/3. Result: 61/12 − (−1/3) = 61/12 + 4/12 = 65/12.
Q: If f(x) = 4x on [1, 5], what is the average value of f?
A: f_avg = (1/(5 − 1)) · ∫ from 1 to 5 of 4x dx = (1/4) · [2x²] from 1 to 5 = (1/4)(50 − 2) = 48/4 = 12.
This material connects directly to u-substitution (Section 5.5), which is the integration counterpart of the chain rule you use here for FTC1. It also sets up every application of integration that follows: area between curves, volumes of revolution, and work all rely on evaluating definite integrals via FTC2. The average value formula reappears in probability and statistics as the expected value of a continuous distribution.
Fundamental Theorem of Calculus, FTC, FTC Part 1, FTC Part 2, Evaluation Theorem, definite integral, antiderivative, chain rule with integrals, variable limits of integration, differentiating an integral, net signed area, average value of a function, MTH 132, Calculus I, Michigan State University, MSU, Sections 4.3 4.4 5.5, Recitation 13, sec² integral, trig antiderivatives, power rule for integration, negative exponents integration