Difficulty: Intermediate | Prerequisites: Calculus I (derivatives, basic antiderivatives), familiarity with sigma notation.
This material sits at the bridge between learning how to differentiate and learning how to integrate. You are moving from Calc I territory (limits, derivatives, basic antiderivatives) into the machinery that makes integration rigorous and practical. Sigma notation and Riemann sums formalise the idea of adding up infinitely many tiny pieces, which is what a definite integral does. The Fundamental Theorem of Calculus (FTC) then ties differentiation and integration together, giving you a shortcut that replaces laborious summation with antiderivative evaluation. If you are comfortable with power-rule differentiation and basic antiderivative formulas, you are ready for this material.
You need three skills for this unit: evaluating closed-form sums using standard sigma formulas, reading a definite integral off a graph by computing net signed area, and applying the Fundamental Theorem of Calculus to both evaluate integrals and differentiate integral-defined functions. Master the formulas, understand what "net area" means (area below the x-axis counts as negative), and know which part of the FTC to reach for.
Sigma notation (summation notation)
A compact way of writing a sum of terms that follow a pattern. The expression ∑(i=1 to n) a_i means "add up a_i for every integer i from 1 to n." Think of it as a for-loop in maths: it tells you the starting value, the ending value, and what to compute at each step.
Definite integral
The definite integral ∫(a to b) f(x) dx represents the net signed area between the curve y = f(x) and the x-axis, from x = a to x = b. In simple terms, it is the total accumulated quantity, where regions below the x-axis subtract from the total.
Net signed area
The result of counting area above the x-axis as positive and area below the x-axis as negative. This is what a definite integral computes, as distinct from "total area" (which would take absolute values).
Antiderivative
A function F(x) whose derivative is f(x). If F'(x) = f(x), then F is an antiderivative of f. In simple terms, it is the reverse of differentiation.
Fundamental Theorem of Calculus, Part 1 (FTC1)
If f is continuous on [a, b], then the function g(x) = ∫(a to x) f(t) dt is differentiable, and g'(x) = f(x). Think of it as: the derivative of a running integral just gives you back the integrand, evaluated at the upper limit.
Fundamental Theorem of Calculus, Part 2 (FTC2, the evaluation theorem)
If F is any antiderivative of f on [a, b], then ∫(a to b) f(x) dx = F(b) - F(a). In simple terms, to evaluate a definite integral, find any antiderivative, plug in the top limit, plug in the bottom limit, and subtract.
Riemann sum
An approximation of a definite integral formed by dividing the interval into subintervals and summing the areas of rectangles. The definite integral is the limit of Riemann sums as the number of rectangles approaches infinity.
The key idea is that you do not need to add terms one by one. Standard closed-form formulas let you jump straight to the answer.
Linearity of summation: You can split a sum across addition and pull constants out.
∑(i=1 to n) [c · a_i + d · b_i] = c · ∑ a_i + d · ∑ b_i
Standard formulas to memorise:
∑(i=1 to n) 1 = n
∑(i=1 to n) i = n(n+1)/2
∑(i=1 to n) i² = n(n+1)(2n+1)/6
∑(i=1 to n) i³ = [n(n+1)/2]²
Worked example (Quiz Problem 1): Find ∑(i=1 to 10) (2i² - 3i + 1).
Split: 2·∑i² - 3·∑i + ∑1
∑(i=1 to 10) i² = 10·11·21/6 = 385
∑(i=1 to 10) i = 10·11/2 = 55
∑(i=1 to 10) 1 = 10
Result: 2(385) - 3(55) + 10 = 770 - 165 + 10 = 615
When you are given a graph of f(x) and asked to evaluate ∫(a to b) f(x) dx, you are computing net signed area. No antiderivative is needed; you read shapes off the picture.
Above the x-axis: area counts as positive.
Below the x-axis: area counts as negative.
Useful geometric shapes: triangles (½ · base · height), rectangles (base · height), trapezoids (½ · (b₁ + b₂) · h), semicircles (½πr²).
Strategy:
Break the interval into sub-regions where the curve forms a recognisable shape.
Compute each region's area with the appropriate geometry formula.
Assign a sign: positive if the region sits above the x-axis, negative if below.
Add them all up.
Worked approach (Quiz Problem 2): For ∫(2 to 9) f(x) dx with the given graph, identify each geometric region between x = 2 and x = 9. The curve appears to form a semicircle above the axis (roughly x = 1 to x = 5, with peak near y = 2), then drops below the axis. Compute each piece, apply signs, and sum. Read coordinates carefully from the grid.
FTC1 tells you how to differentiate a function defined as an integral with a variable limit.
Standard form: If g(x) = ∫(a to x) f(t) dt, then g'(x) = f(x).
Variable in the lower limit: If the variable is in the lower limit instead, flip the sign.
If g(x) = ∫(x to b) f(t) dt, rewrite as g(x) = -∫(b to x) f(t) dt, so g'(x) = -f(x).
Chain rule variant: If the upper limit is a function u(x) rather than plain x:
g(x) = ∫(a to u(x)) f(t) dt implies g'(x) = f(u(x)) · u'(x).
Worked example (Quiz Problem 3): For f(x) = ∫(x to 4) sin(t²) dt, find f'(x).
Rewrite: f(x) = -∫(4 to x) sin(t²) dt.
By FTC1: the derivative of ∫(4 to x) sin(t²) dt is sin(x²).
Therefore f'(x) = -sin(x²).
Key point: You do not need to find an antiderivative of sin(t²). In fact, sin(t²) has no elementary antiderivative. FTC1 sidesteps that entirely.
This is the bread-and-butter computation: find an antiderivative, evaluate at the endpoints, subtract.
Process:
Find F(x), any antiderivative of the integrand f(x).
Compute F(b) - F(a).
Common antiderivatives you need:
∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, provided n ≠ -1
∫ sin(x) dx = -cos(x) + C
∫ cos(x) dx = sin(x) + C
∫ eˣ dx = eˣ + C
∫ 1/x dx = ln|x| + C
Worked example (Quiz Problem 4): Calculate ∫(1 to 2) (6x² - 2x + 7) dx.
Antiderivative: F(x) = 2x³ - x² + 7x.
F(2) = 2(8) - 4 + 14 = 16 - 4 + 14 = 26.
F(1) = 2(1) - 1 + 7 = 8.
Result: 26 - 8 = 18.
No "+ C" needed in definite integrals. The constant cancels when you subtract F(b) - F(a).
Summation formulas:
∑(i=1 to n) 1 = n
∑(i=1 to n) i = n(n+1)/2
∑(i=1 to n) i² = n(n+1)(2n+1)/6
∑(i=1 to n) i³ = [n(n+1)/2]²
FTC1 (derivative of an integral):
d/dx ∫(a to x) f(t) dt = f(x)
d/dx ∫(a to u(x)) f(t) dt = f(u(x)) · u'(x)
d/dx ∫(x to b) f(t) dt = -f(x)
FTC2 (evaluation theorem):
∫(a to b) f(x) dx = F(b) - F(a), where F'(x) = f(x)
Power rule for antiderivatives:
∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, for n ≠ -1
Forgetting the sign flip when the variable is in the lower limit. If f(x) = ∫(x to 4) g(t) dt, then f'(x) = -g(x), not +g(x). Students routinely drop the negative.
Confusing net signed area with total area. A definite integral gives net area (below-axis regions subtract). If a problem asks for "total area," you must integrate |f(x)|, splitting at the zeros. The quiz asks for the integral, so net signed area is what you want.
Adding "+ C" to a definite integral. The constant of integration is only needed for indefinite integrals. In a definite integral, the C cancels.
Trying to find an antiderivative when FTC1 is the right tool. Problem 3 asks for the derivative of an integral-defined function. You do not need (and cannot find) an elementary antiderivative of sin(t²). FTC1 gives you the answer directly.
⚠️ Summation formula questions appear on nearly every Calc II midterm. Memorise all four standard formulas and practise splitting sums by linearity.
⚠️ Graph-based integral questions test whether you understand the geometric meaning of the integral, not just mechanical antidifferentiation. Read the grid carefully and identify shapes before computing.
⚠️ FTC1 with a variable lower limit (or a composed upper limit) is a classic exam trap. Always check which limit has the variable, and apply the chain rule if needed.
⚠️ A straightforward "evaluate the definite integral" problem is free marks if you know your antiderivative rules. Practise until the power rule and basic trig antiderivatives are automatic.
True or false: ∑(i=1 to n) i² = n(n+1)/2. (False, that is the formula for ∑i. The correct formula is n(n+1)(2n+1)/6.)
True or false: ∫(a to b) f(x) dx can be negative. (True, when the region sits mostly below the x-axis.)
Fill in the blank: d/dx ∫(x to 5) cos(t) dt = ____. (-cos(x))
True or false: You need to add "+ C" when evaluating ∫(1 to 3) x² dx. (False, constants cancel in definite integrals.)
Fill in the blank: ∫(1 to 2) 4x³ dx = ____. (F(x) = x⁴, so 2⁴ - 1⁴ = 16 - 1 = 15.)
Q: Evaluate ∑(i=1 to 5) (3i² + 2).
A: Split into 3·∑i² + ∑2. We have ∑(i=1 to 5) i² = 5·6·11/6 = 55. And ∑(i=1 to 5) 2 = 10. So the answer is 3(55) + 10 = 175.
Q: If g(x) = ∫(2 to x³) ln(t) dt, what is g'(x)?
A: By FTC1 with the chain rule: g'(x) = ln(x³) · 3x² = 3x² · 3ln(x) = 9x² ln(x). Equivalently, ln(x³) = 3ln(x), so g'(x) = 3x² · 3ln(x) = 9x²ln(x).
Q: Evaluate ∫(0 to 3) (2x + 1) dx using FTC2.
A: Antiderivative: F(x) = x² + x. Then F(3) - F(0) = (9 + 3) - 0 = 12.
Q: The graph of f(x) forms a triangle with vertices at (0, 0), (4, 0), and (4, 6). What is ∫(0 to 4) f(x) dx?
A: The triangle sits above the x-axis with base 4 and height 6. Area = ½ · 4 · 6 = 12. Since it is above the axis, the integral is +12.
Q: If h(x) = ∫(x² to 7) e^t dt, find h'(x).
A: Rewrite as h(x) = -∫(7 to x²) e^t dt. By FTC1 + chain rule: h'(x) = -(e^(x²)) · 2x = -2x · e^(x²).
Summation and Riemann sums connect directly to the formal definition of the definite integral, which you will revisit when studying improper integrals and convergence of series later in the course. The Fundamental Theorem of Calculus is the backbone of every integration technique you will learn next (substitution, integration by parts, partial fractions), because each of those methods produces an antiderivative that you then evaluate at endpoints. The geometric interpretation of integrals also resurfaces when computing areas between curves, volumes of revolution, and arc length.
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