Difficulty: Intermediate | Prerequisites: Sections 13.1 and 13.2 (indefinite integrals and substitution).
Definite integrals take integration from the abstract ("find the family of antiderivatives") to the concrete ("compute a number"). The Fundamental Theorem of Calculus connects the antiderivative you found in Section 13.1 to the area under a curve between two bounds. Riemann sums approximate that area using rectangles, which is how the definite integral is defined and why numerical estimates work. Section 13.4 applies these tools to real problems: displacement from velocity, revenue from marginal revenue, accumulated quantity from a rate.
To evaluate a definite integral, find the antiderivative, plug in the upper bound, plug in the lower bound, and subtract. Riemann sums approximate the integral by summing rectangle areas. Applications include displacement, total revenue, total quantity, and area under a curve.
Definite integral
The integral ∫ from a to b of f(x) dx, which evaluates to a specific number representing the net signed area between f(x) and the x-axis on [a, b]. Think of it as the indefinite integral with boundaries that turn the family of functions into a single value.
Fundamental Theorem of Calculus (FTC)
If F(x) is an antiderivative of f(x), then ∫ from a to b of f(x) dx = F(b) − F(a). In simple terms, find the antiderivative, evaluate it at both bounds, and subtract.
Left Riemann sum
An approximation of a definite integral that divides [a, b] into n equal subintervals and uses the function value at the left endpoint of each subinterval as the rectangle height. Think of it as stacking n rectangles under the curve, each touching the curve at its left edge.
Subinterval width (Δx)
The width of each rectangle in a Riemann sum, calculated as (b − a)/n. All subintervals have the same width in these problems.
Displacement
The net change in position, found by integrating velocity over a time interval: ∫ from a to b of v(t) dt. In simple terms, it tells you how far the object ended up from where it started, accounting for direction.
Net change
The total accumulation of a rate over an interval. Integrating a rate function (marginal revenue, upload rate, velocity) gives the total amount gained or lost over that interval.
The process is always the same:
Find the antiderivative F(x) (no + C needed for definite integrals)
Evaluate F(b) − F(a)
Simplify to get an exact answer
Example: ∫ from 0 to 1 of (6x⁵ + 15x⁴ − 9x² + 1) dx. The antiderivative is x⁶ + 3x⁵ − 3x³ + x. Evaluate at 1: 1 + 3 − 3 + 1 = 2. Evaluate at 0: 0. Answer: 2.
When the integrand requires substitution, you have two options:
Option A: find the antiderivative using substitution (as in Section 13.2), substitute back to x, then evaluate at the bounds
Option B: change the limits of integration to match u. If u = g(x), the new lower bound is g(a) and the new upper bound is g(b). Integrate in terms of u with the new bounds and never substitute back
Example: ∫ from ln 3 to ln 5 of e^(2x) dx. Antiderivative is (1/2)e^(2x). Evaluate: (1/2)e^(2 ln 5) − (1/2)e^(2 ln 3) = (1/2)(25) − (1/2)(9) = 8.
Example: ∫ from e³ to e⁵ of 2/x dx = 2 ln|x| evaluated from e³ to e⁵ = 2(5) − 2(3) = 4.
To compute a left Riemann sum with n subintervals on [a, b]:
Calculate Δx = (b − a)/n
List the left endpoints: x₀ = a, x₁ = a + Δx, x₂ = a + 2Δx, ..., x_{n−1} = a + (n − 1)Δx
Evaluate f at each left endpoint
Sum: Δx × [f(x₀) + f(x₁) + ... + f(x_{n−1})]
Example: f(x) = 3x² + 2x − 3 on [1, 3] with n = 5. Δx = (3 − 1)/5 = 0.4. Left endpoints: 1, 1.4, 1.8, 2.2, 2.6. Evaluate f at each, multiply each by 0.4, and sum.
Displacement from velocity
Displacement = ∫ from a to b of v(t) dt. This gives the net change in position (positive minus negative movement).
Total revenue from marginal revenue
Revenue from selling units a through b = ∫ from a to b of MR(x) dx.
Total accumulated quantity from a rate
If f(t) is a rate (million hours of video per year), then ∫ from a to b of f(t) dt gives the total accumulated over that period.
Area under a curve
When f(x) ≥ 0 on [a, b], the area between f(x) and the x-axis is simply ∫ from a to b of f(x) dx.
\int_a^b f(x)\,dx = F(b) - F(a)\Delta x = \frac{b-a}{n}\text{Left Riemann sum} = \Delta x \sum_{i=0}^{n-1} f(x_i)\text{Displacement} = \int_a^b v(t)\,dtDefinite integrals are how engineers and scientists compute totals from rates. Integrating a flow rate gives total volume; integrating power gives total energy consumed; integrating a marginal revenue function gives total revenue over a range of sales.
Students add + C to definite integrals. You do not. The constant cancels when you subtract F(a) from F(b).
Students confuse displacement with distance. Displacement is the net integral (which can be negative); total distance requires integrating the absolute value of velocity.
Students use the right endpoints instead of the left endpoints in a left Riemann sum. In a left sum, you never use f(b) as a rectangle height.
Students forget to use the correct units in applied problems. If velocity is in metres per second and time is in seconds, displacement is in metres.
⚠️ "Give the exact value" means leave your answer as a fraction, logarithm, or expression with e. Do not convert to a decimal unless asked.
⚠️ When evaluating e^(ln something), simplify: e^(ln k) = k and e^(2 ln k) = k².
⚠️ For Riemann sum problems, write out every rectangle's height and width explicitly. The exam asks you to show all work.
⚠️ Applied problems: read the bounds carefully. "Selling box 101 through 5,000" means the integral runs from 100 to 5,000 (or 101 to 5,000, depending on convention; check how the problem sets up the variable).
True or false: ∫ from 0 to 1 of 3x² dx = x³ evaluated from 0 to 1 = 1. (True.)
Fill in the blank: for a left Riemann sum with n = 4 on [2, 3], Δx = ____. (0.25)
True or false: displacement can be negative. (True. If the object moves backwards more than forwards, net displacement is negative.)
Fill in the blank: e^(2 ln 5) simplifies to ____. (25)
True or false: ∫ from a to b of f(x) dx always needs + C. (False. Definite integrals evaluate to a number.)
Q: Evaluate ∫ from 0 to 1 of (6x⁵ + 15x⁴ − 9x² + 1) dx.
A: Antiderivative: x⁶ + 3x⁵ − 3x³ + x. At x = 1: 1 + 3 − 3 + 1 = 2. At x = 0: 0. Answer: 2.
Q: Evaluate ∫ from 2 to 7 of (x + 5/x) dx.
A: Antiderivative: (1/2)x² + 5 ln|x|. At x = 7: 49/2 + 5 ln 7. At x = 2: 2 + 5 ln 2. Answer: 45/2 + 5 ln(7/2).
Q: Evaluate ∫ from 0 to 6 of e^(−x + 6) dx.
A: Antiderivative: −e^(−x + 6). At x = 6: −e⁰ = −1. At x = 0: −e⁶. Answer: −1 − (−e⁶) = e⁶ − 1.
Q: Evaluate ∫ from ln 3 to ln 5 of e^(2x) dx.
A: Antiderivative: (1/2)e^(2x). At x = ln 5: (1/2)e^(2 ln 5) = (1/2)(25) = 25/2. At x = ln 3: (1/2)(9) = 9/2. Answer: 25/2 − 9/2 = 8.
Q: Solve ∫ from 2 to b of (2x − 4) dx = 9 for positive b.
A: Antiderivative: x² − 4x. Evaluate: (b² − 4b) − (4 − 8) = b² − 4b + 4 = (b − 2)² = 9. So b − 2 = 3 (positive), b = 5.
Q: Find the displacement of a particle with v(t) = −t² + 8 from t = 2 to t = 6.
A: ∫ from 2 to 6 of (−t² + 8) dt. Antiderivative: −(1/3)t³ + 8t. At t = 6: −72 + 48 = −24. At t = 2: −8/3 + 16 = 40/3. Answer: −24 − 40/3 = −72/3 − 40/3 = −112/3 metres.
Q: Find the revenue from selling box 101 through 5,000 if marginal revenue is 500e^(−0.001x).
A: ∫ from 100 to 5000 of 500e^(−0.001x) dx. Antiderivative: 500/(−0.001) e^(−0.001x) = −500,000 e^(−0.001x). Evaluate: −500,000 e^(−5) − (−500,000 e^(−0.1)) = 500,000(e^(−0.1) − e^(−5)).
Q: Calculate the area bounded by y = √x, the x-axis, x = 0 and x = 16.
A: ∫ from 0 to 16 of x^(1/2) dx = (2/3)x^(3/2) evaluated from 0 to 16 = (2/3)(64) − 0 = 128/3.
Definite integrals rely on the antiderivative techniques from Sections 13.1 and 13.2. The Riemann sum concept reappears in numerical methods and statistics courses. The area-under-a-curve interpretation leads directly into Section 14.2 (area between curves), where you integrate the difference of two functions rather than a single function.
definite integral, Fundamental Theorem of Calculus, FTC, left Riemann sum, Riemann approximation, displacement, net change, area under a curve, marginal revenue integration, accumulated quantity, MAT 210, brief calculus, Sections 13.3, 13.4, ASU calculus