Difficulty: Intermediate | Prerequisites: Differentiation rules (Ch. 3), basic antiderivatives, Riemann sums concept.
This topic cluster covers the machinery that connects Riemann sums to the elegant evaluation shortcut most of calculus relies on. You learn what a definite integral is (a limit of sums), why the Fundamental Theorem lets you skip the limit and just plug in antiderivative values, and how the Net Change Theorem turns rate-of-change information into cumulative totals. Master these and every later integration technique is just a way of finding the antiderivative that FTC2 asks for.
Definite integral
The limit of Riemann sums as the number of subintervals approaches infinity: for f continuous on [a, b], the definite integral from a to b is the signed area between the curve and the x-axis.
In simple terms, this means: chop the region under a curve into thin rectangles, add their areas, and take the limit as they get infinitely thin.
Indefinite integral (antiderivative)
A family of functions F(x) + C whose derivative equals the integrand: if F'(x) = f(x), then the indefinite integral of f(x) dx = F(x) + C.
Think of it as: the reverse of differentiation, but you always tack on + C because a whole family of curves shares the same slope function.
Riemann sum
A finite approximation of a definite integral formed by summing f(x_i) times the subinterval width across n rectangles.
In simple terms, this means: pick sample heights along the curve, multiply each by the strip width, and add them up.
Integrability
A function is integrable on [a, b] if the definite integral exists. Continuous functions and functions with only finitely many jump discontinuities are integrable.
Think of it as: if the function does not misbehave too wildly, you can compute its total signed area.
Fundamental Theorem of Calculus, Part 1 (FTC1)
If f is continuous on [a, b], then g(x) = integral from a to x of f(t) dt is differentiable and g'(x) = f(x). Differentiation undoes integration.
In simple terms, this means: the rate of change of the accumulated area under f, at any point x, is just the height f(x).
Fundamental Theorem of Calculus, Part 2 (FTC2)
If f is continuous on [a, b] and F is any antiderivative of f, then the definite integral from a to b of f(x) dx = F(b) - F(a).
Think of it as: instead of doing a painful limit of sums, just find an antiderivative, plug in the endpoints, and subtract.
Net Change Theorem
The integral of a rate of change over [a, b] gives the net change in the original quantity: integral from a to b of F'(x) dx = F(b) - F(a). Equivalently, F(b) = F(a) + integral of the rate.
In simple terms, this means: if you know how fast something is changing, integrating that rate tells you the total change.
Constant of integration (+ C)
The arbitrary constant appended to every indefinite integral, reflecting the fact that infinitely many antiderivatives exist (each shifted vertically). Definite integrals do not carry + C because the constant cancels in the subtraction F(b) - F(a).
Given f(x) continuous on [a, b], divide the interval into n equal subintervals of width delta-x. Choose a sample point x_i in each subinterval. The definite integral is:
\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\,\Delta xThe function f need not be continuous everywhere: if f has only finitely many jump discontinuities on [a, b], the integral still exists.
These properties let you break apart, recombine, and bound integrals without evaluating them.
Reversing limits: Integral from a to b of f(x) dx = negative of integral from b to a of f(x) dx. Swapping the limits flips the sign.
Zero-width interval: Integral from a to a of f(x) dx = 0. No width, no area.
Interval addition: Integral from a to b = integral from a to c + integral from c to b. You can split an integral at any interior point c.
Linearity (sum/difference): Integral of [f(x) +/- g(x)] dx = integral of f(x) dx +/- integral of g(x) dx. Integration distributes over addition and subtraction.
Linearity (constant multiple): Integral of C times f(x) dx = C times integral of f(x) dx. Constants slide out of the integral.
Comparison: If f(x) <= g(x) on [a, b], then integral of f <= integral of g. Bigger function, bigger integral.
Bounding: If m <= f(x) <= M on [a, b], then m(b - a) <= integral of f(x) dx <= M(b - a). The integral is trapped between the areas of the smallest and largest bounding rectangles.
If f is continuous on [a, b], define g(x) = integral from a to x of f(t) dt. Then g is continuous on [a, b], differentiable on (a, b), and:
g'(x) = f(x)The key insight: differentiation and integration are inverse operations. The derivative of the "area so far" function is just the integrand evaluated at the upper limit.
When the upper limit is not simply x but some function of x, apply the chain rule. For example, if g(x) = integral from 1 to x^3 of sec(t^2) dt, then g'(x) = sec((x^3)^2) times 3x^2.
When both limits are functions of x, split into two integrals using a constant and differentiate each piece.
If f is continuous on [a, b] and F is any antiderivative of f (i.e., F' = f), then:
\int_a^b f(x)\,dx = F(b) - F(a)This converts the problem of computing a definite integral from a limit of sums into a straightforward evaluation: find F, plug in the endpoints, subtract. Every integration technique you learn from here on is ultimately a method for finding that antiderivative F.
The notation integral of f(x) dx = F(x) + C means F'(x) = f(x). The "+C" is essential for indefinite integrals because differentiation destroys constants. (For definite integrals the constant cancels, so it does not appear.)
The FTC2 can be restated in a physically intuitive way. If F'(x) is the rate of change of some quantity F, then:
\int_a^b F'(x)\,dx = F(b) - F(a)Equivalently: the new value = initial value + integral of the rate of change.
Applications follow directly:
If v(t) is velocity, the integral of v(t) from t1 to t2 gives net displacement (signed). For total distance, integrate |v(t)|.
If V'(t) is the rate water flows into a reservoir, the integral gives the net change in volume.
If g(t) is a population growth rate, the integral gives the population increase over that period.
Function | Antiderivative |
|---|---|
k (constant) | kx + C |
x^n (n not equal to -1) | x^(n+1) / (n+1) + C |
1/x | ln |
e^x | e^x + C |
b^x | b^x / ln(b) + C |
cos x | sin x + C |
sin x | -cos x + C |
sec^2 x | tan x + C |
csc^2 x | -cot x + C |
sec x tan x | sec x + C |
csc x cot x | -csc x + C |
1 / sqrt(1 - x^2) | arcsin x + C |
1 / (1 + x^2) | arctan x + C |
sinh x | cosh x + C |
cosh x | sinh x + C |
Linearity rules to remember:
Integral of [f(x) +/- g(x)] dx = integral of f(x) dx +/- integral of g(x) dx
Integral of K times f(x) dx = K times integral of f(x) dx
The Net Change Theorem is the reason engineers can compute how much water a reservoir has gained from flow-rate data, or how far a particle has travelled from its velocity profile. In population biology, if you know the growth rate g(t) of a population and its initial size, the integral of g(t) gives the total increase, so the new population is the old population plus that integral.
Students often think that the integral of a product equals the product of the integrals: integral of f(x) g(x) dx = integral of f(x) dx times integral of g(x) dx. This is false. Integration has no product rule in that sense.
Forgetting +C on indefinite integrals is the single most common slip. The "integration by parts paradox" (where applying parts to 1/x appears to prove 0 = 1) exists precisely because +C was dropped from both sides.
Confusing displacement with total distance travelled. The integral of velocity gives net displacement (signed). To get total distance, you must integrate the absolute value of velocity, which typically means splitting the integral wherever v(t) changes sign.
Assuming FTC2 applies when the integrand is not continuous on the interval. If f has an infinite discontinuity inside [a, b], you need improper integral techniques instead.
FTC1 with chain rule (e.g. d/dx of integral from 1 to x^3 of f(t) dt) is a favourite exam question. Remember to multiply by the derivative of the upper limit.
Expect at least one question asking you to distinguish displacement from distance. Know when to split the integral.
The seven properties of integrals frequently appear as "simplify this combination of integrals into a single integral" problems (e.g. combining integrals with overlapping limits via interval addition).
The bounding/comparison property is tested as "show that the integral lies between two values" without evaluating the integral itself.
Net change word problems (reservoirs filling, populations growing, particles moving) are standard. Read the problem for initial value, rate function, and time interval.
True or False: The definite integral from a to a of any continuous function is always 0. (True.)
Fill in the blank: If F'(x) = f(x), then integral from a to b of f(x) dx = ______. (F(b) - F(a).)
True or False: Integral of f(x) g(x) dx = (integral of f(x) dx)(integral of g(x) dx). (False.)
Fill in the blank: For indefinite integrals, you must always include ______ at the end. (+ C.)
True or False: The integral of velocity from t = 0 to t = 5 always equals the total distance travelled. (False, it gives net displacement; total distance requires integrating |v(t)|.)
Q: Write the combination integral from -2 to 2 of f(x) dx + integral from 2 to 5 of f(x) dx - integral from -2 to -1 of f(x) dx as a single integral of the form integral from a to b of f(x) dx.
A: Integral from -1 to 5 of f(x) dx. Use interval addition to merge the first two, then subtract the portion from -2 to -1.
Q: A fish population in a lake grows at a rate of f(t) = 10 + 5t^(3/2) fish per year. How much does the population increase in 4 years?
A: Integrate f(t) from 0 to 4. The antiderivative is 10t + 5 times (2/5) t^(5/2) = 10t + 2t^(5/2). Evaluating: 10(4) + 2(4)^(5/2) - 0 = 40 + 2(32) = 104 fish.
Q: A particle has velocity v(t) = 3t - 5 for 0 <= t <= 3. Find (a) net displacement and (b) total distance travelled.
A: (a) Integral from 0 to 3 of (3t - 5) dt = [3t^2/2 - 5t] from 0 to 3 = (27/2 - 15) - 0 = -3/2 m. (b) v(t) = 0 at t = 5/3. Split: integral from 0 to 5/3 of |3t - 5| dt + integral from 5/3 to 3 of |3t - 5| dt. The total distance is 41/6 m.
Q: Use the comparison property to show that 2 <= integral from -1 to 1 of sqrt(1 + x^2) dx <= 2 sqrt(2).
A: On [-1, 1], x^2 ranges from 0 to 1, so 1 <= 1 + x^2 <= 2, giving 1 <= sqrt(1 + x^2) <= sqrt(2). Multiply through by (b - a) = 2: 2 <= integral <= 2 sqrt(2).
Q: A tank with capacity 1000 litres fills at rate R(t) = 60 - t litres per minute. How long until it is full?
A: Solve integral from 0 to T of (60 - t) dt = 1000. That gives 60T - T^2/2 = 1000, so T^2 - 120T + 2000 = 0, yielding T = 20 minutes (taking the smaller root).
The definite integral reappears in Chapter 6 (applications of integration: areas between curves, volumes of revolution, arc length) where you set up the integral and then use Chapter 7 techniques to evaluate it. The Net Change Theorem is the one-dimensional precursor to the line integral and the divergence theorem in multivariable calculus. Riemann sums return in Section 7.7 as the basis for numerical approximation methods (trapezoidal rule, Simpson's rule).
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