Difficulty: Intermediate | Prerequisites: Riemann sums study notes (Doc 1), basic area formulas (triangles, rectangles, semicircles).
Once you can approximate integrals with Riemann sums, the next step is evaluating them exactly, either from a graph (using geometry) or by applying the algebraic properties of integrals. This set of skills also introduces the distinction between displacement and total distance, a difference that catches many students off guard on exams. These ideas feed directly into the Fundamental Theorem of Calculus.
You can evaluate definite integrals by computing geometric areas from graphs (watch the sign below the x-axis). Displacement is the integral of velocity; total distance is the integral of the absolute value of velocity. The linearity and interval-splitting properties let you break complex integrals into simpler pieces you already know.
Displacement
The net change in position. Computed as the integral of velocity. If an object moves forward 10 feet then backward 3 feet, its displacement is 7 feet. Areas below the x-axis count as negative.
Total distance travelled
The total length of the path, regardless of direction. Computed as the integral of the absolute value of velocity. In the same example, the total distance is 13 feet. Areas below the x-axis are made positive before summing.
Net signed area
The value of a definite integral interpreted geometrically: area above the x-axis is positive, area below is negative. The integral gives the net result, not the total area.
Linearity of the integral
The property that the integral of a sum equals the sum of the integrals, and constants can be pulled out. Formally: the integral of [af(x) + bg(x)] dx = a times the integral of f(x) dx + b times the integral of g(x) dx.
Interval additivity (splitting property)
The integral from a to c of f(x) dx = the integral from a to b of f(x) dx + the integral from b to c of f(x) dx for any value b between a and c. Think of it as: you can split the journey into legs and add them.
Reversing limits of integration
The integral from a to b of f(x) dx = negative the integral from b to a of f(x) dx. Swapping the limits flips the sign.
When a graph of v(t) or f(x) is made of straight-line segments, you evaluate the definite integral by computing the areas of the geometric shapes formed between the curve and the x-axis.
Common shapes and their areas:
Triangle: (1/2)(base)(height)
Rectangle: (base)(height)
Trapezoid: (1/2)(sum of parallel sides)(height)
Semicircle: (1/2)(pi)(r^2)
Critical rule: regions above the x-axis contribute positive area; regions below contribute negative area.
Given a piecewise-linear velocity graph v(t) on [0, 16], the integral from 0 to 16 of v(t) dt is computed by summing the signed areas of each geometric piece.
From the practice set: (1/2)(9 + 3)(4) - (1/2)(3)(1) + (1/2)(4)(2) = 24 - 3/2 + 4 = 26.5 feet.
Interpretation: between t = 0 and t = 16 seconds, the object travelled a net 26.5 feet from its original location. The word "net" matters because some of the motion was in the negative direction.
To find total distance, integrate the absolute value of velocity: the integral from 0 to 16 of |v(t)| dt.
Every piece of area becomes positive, including the region below the x-axis:
(1/2)(9 + 3)(4) + (1/2)(3)(1) + (1/2)(4)(2) = 24 + 3/2 + 4 = 29.5 feet.
Interpretation: the object covered 29.5 feet of total ground between t = 0 and t = 16 seconds.
If v(t) changes sign on the interval, displacement and total distance will differ. Displacement uses signed area (the integral of v). Total distance uses unsigned area (the integral of |v|). If v(t) never goes negative, the two are equal.
You can factor out constants and split sums:
Integral of [2f(x) - 3h(x)] dx = 2 * integral of f(x) dx - 3 * integral of h(x) dx
Worked example: given the integral from 7 to 9 of f(x) dx = 5 and the integral from 7 to 9 of h(x) dx = 4, find the integral from 7 to 9 of [2f(x) - 3h(x)] dx.
Answer: 2(5) - 3(4) = 10 - 12 = -2.
The integral from a to c of f(x) dx = the integral from a to b of f(x) dx + the integral from b to c of f(x) dx.
Worked example: given the integral from 1 to 9 of f(x) dx = -1 and the integral from 7 to 9 of f(x) dx = 5, find the integral from 1 to 7 of f(x) dx.
Since the integral from 1 to 7 + the integral from 7 to 9 = the integral from 1 to 9, we get the integral from 1 to 7 of f(x) dx + 5 = -1, so the integral from 1 to 7 of f(x) dx = -6.
To find the integral from 7 to 1, reverse the limits: the integral from 7 to 1 of f(x) dx = -(-6) = 6.
Worked example: given that the integral from -2 to 8 of (3f(x) + 2) dx = 35 and the integral from 5 to -2 of f(x) dx = -12.
First, note that the integral from 5 to -2 of f(x) dx = -12 means the integral from -2 to 5 of f(x) dx = 12 (reverse limits, flip sign).
Expand: 3 * integral from -2 to 8 of f(x) dx + integral from -2 to 8 of 2 dx = 35.
The integral from -2 to 8 of 2 dx = 2(10) = 20.
So 3 * integral from -2 to 8 of f(x) dx = 15, giving the integral from -2 to 8 of f(x) dx = 5.
Split: the integral from -2 to 5 + the integral from 5 to 8 = 5. Since the integral from -2 to 5 = 12, the integral from 5 to 8 of f(x) dx = 5 - 12 = -7.
The integral from a to b of f'(x) dx = f(b) - f(a). This is the Fundamental Theorem of Calculus, Part 1. It also means that the integral from 0 to 9 of f'(x) dx equals f(9) - f(0).
But you can also evaluate the integral from 0 to 9 of f'(x) dx directly from the graph of f'(x) by computing signed geometric areas, just as you would with any other function's graph.
Worked example from the practice set: the graph of f'(x) on [0, 9] consists of line segments and a semicircle. Breaking it into geometric pieces:
[0, 1]: triangle with base 1, height 2 = area (1/2)(1)(2) = 1
[1, 2]: rectangle with base 1, height 2 = area 2
[2, 3]: rectangle with base 1, height 2 = area 2
[3, 5]: semicircle below the axis with radius 1 = area -(1/2)(pi)(1)^2 = -pi/2
[5, 7]: triangle with base 2, height 3/2 = area (1/2)(2)(3/2) = 3/2 (need to check exact shape)
[7, 9]: triangle below the axis
The result: 1 + 2 + 2 - pi/2 + 3/4 - 3/4 = 5 - pi/2.
Since the integral from 0 to 9 of f'(x) dx = f(9) - f(0), if you know f(9) = 1:
f(9) - f(0) = 5 - pi/2, so 1 - f(0) = 5 - pi/2, giving f(0) = 1 - (5 - pi/2) = -4 + pi/2.
Alternatively, think of it as f(0) = f(9) - the integral from 0 to 9 of f'(x) dx = 1 - (5 - pi/2) = -4 + pi/2.
This is a very common exam pattern: you are given the graph of f', asked to compute an integral of f' geometrically, and then use the FTC relationship to find f at a different point.
For a particle with velocity v(t) = sin(t) starting at x = 5, the total distance from t = 0 to t = 3pi is the integral from 0 to 3pi of |sin(t)| dt.
The initial position x = 5 is irrelevant to distance. Distance depends only on the integral of |v|, not on where the particle started. This is why the answer is option (C), not option (D).
Displacement
Displacement = integral from a to b of v(t) dt
Total distance
Total distance = integral from a to b of |v(t)| dt
FTC relationship (recovering function values)
f(b) = f(a) + integral from a to b of f'(x) dx
Linearity
Integral of [af(x) + bg(x)] dx = a * integral of f(x) dx + b * integral of g(x) dx
Interval splitting
Integral from a to c = integral from a to b + integral from b to c
Reversing limits
Integral from a to b = -(integral from b to a)
Integral of a constant
Integral from a to b of c dx = c(b - a)
Semicircle area (for graph problems)
(1/2)(pi)(r^2), with sign determined by whether the region is above or below the x-axis.
Displacement vs. total distance matters in navigation. A GPS showing your position after a walk gives displacement; a fitness tracker counting every step gives total distance. In physics, the integral of a force-displacement curve gives work done, and the sign matters: negative work means energy is taken out of the system. In pharmacokinetics, the area under the curve (AUC) of a drug concentration graph determines dosing intervals.
Students treat displacement and total distance as the same thing. They are not. If velocity is ever negative on the interval, displacement will be smaller than total distance.
When computing areas from a graph, students forget to make regions below the x-axis negative for displacement (or positive for total distance). Always check the sign.
Students add the initial position to the total distance integral. Total distance is the integral of |v(t)|, full stop. The starting position x(0) does not appear in it.
When splitting integrals, students forget that reversing the limits changes the sign. The integral from 5 to -2 is the negative of the integral from -2 to 5.
Displacement vs. total distance is a near-certain exam question. Make sure you can articulate the difference in a sentence and know which integral to write.
"Using correct units, explain the meaning of..." is a free-response staple. The grading rubric typically requires: the correct units, a contextual description (not just "the area under the curve"), and a reference to the specific interval.
Properties-of-integrals problems (linearity, splitting) often combine several given integrals into a multi-step calculation. Show every algebraic step.
Graph-of-f' problems test whether you can identify geometric shapes and compute their areas, including semicircles. Know the area formula for a semicircle.
True or false: the integral of |v(t)| gives displacement.
False. The integral of |v(t)| gives total distance. The integral of v(t) (without absolute value) gives displacement.
Fill in the blank: the integral from a to b of f(x) dx = ______ times the integral from b to a of f(x) dx.
-1.
True or false: if the integral from 2 to 8 of f(x) dx = 10 and the integral from 2 to 5 of f(x) dx = 3, then the integral from 5 to 8 of f(x) dx = 7.
True. By the splitting property, 3 + the integral from 5 to 8 = 10.
Fill in the blank: the area of a semicircle with radius r is ______.
(1/2)(pi)(r^2).
Q: A particle moves along the x-axis with velocity v(t). The integral from 0 to 10 of v(t) dt = 15 and the integral from 0 to 10 of |v(t)| dt = 23. Explain what each number tells you.
A: 15 is the displacement, the net change in position. The particle ended up 15 units to the right of where it started. 23 is the total distance, the full length of ground the particle covered, counting all motion regardless of direction.
Q: Given the integral from 1 to 9 of f(x) dx = -1 and the integral from 7 to 9 of f(x) dx = 5, find the integral from 7 to 1 of f(x) dx.
A: First, the integral from 1 to 7 of f(x) dx = (integral from 1 to 9) - (integral from 7 to 9) = -1 - 5 = -6. Then the integral from 7 to 1 of f(x) dx = -(-6) = 6.
Q: The graph of f'(x) on [0, 4] forms a triangle above the x-axis with base 4 and height 6. If f(0) = 3, find f(4).
A: The integral from 0 to 4 of f'(x) dx = (1/2)(4)(6) = 12. By the FTC, f(4) = f(0) + 12 = 3 + 12 = 15.
Q: For v(t) = sin(t), which integral gives the total distance from t = 0 to t = 3pi: the integral of sin(t) dt or the integral of |sin(t)| dt?
A: The integral of |sin(t)| dt. Since sin(t) changes sign on [0, 3pi], the unsigned integral is needed for total distance.
Q: If the integral from -2 to 8 of (3f(x) + 2) dx = 35, what is the integral from -2 to 8 of f(x) dx?
A: Expand using linearity: 3 * integral of f(x) dx + integral of 2 dx = 35. The integral of 2 dx from -2 to 8 is 2(10) = 20. So 3 * integral of f(x) dx = 15, giving the integral from -2 to 8 of f(x) dx = 5.
The displacement vs. total distance distinction reappears in multivariable calculus (vector-valued functions) and physics (kinematics). The integral properties here are the foundation for every integration technique in Calculus II, including substitution and integration by parts. The graph-of-f' problems connect directly to curve sketching and the first/second derivative tests from earlier units.
Definite integral, displacement, total distance, net signed area, absolute value integral, linearity of integrals, integral properties, splitting integrals, reversing limits, interval additivity, geometric area, velocity graph, area under a curve, FTC, Fundamental Theorem of Calculus, f(b) - f(a), semicircle area, piecewise linear, Calculus I, AP Calculus AB, Unit 7, MTH 132, Michigan State University