Difficulty: Intermediate | Prerequisites: Parts 1 and 2 of these notes, antiderivative rules, properties of integrals
Once you understand what the FTC says (Parts 1 and 2 of these notes), the bulk of exam practice is applying it: evaluating definite integrals by hand, computing enclosed areas, working with piecewise functions, recovering function values from derivative information, and reading accumulation functions off a graph. This section covers all of those computational and analytical skills. These are the problems that fill the majority of any FTC worksheet or exam.
Evaluating definite integrals means finding an antiderivative and subtracting endpoint values. Area problems require attention to sign (absolute value when the curve dips below the axis). Graph-based problems test whether you can read accumulated area, identify extrema, and determine concavity of an accumulation function from the graph of its derivative.
Definite integral evaluation (FTC2 in practice)
Find an antiderivative F(x), then compute F(b) - F(a). No "+C" needed because it cancels.
Signed area vs. geometric area
The definite integral gives signed area: regions below the x-axis contribute negatively. To find the geometric (total positive) area, take the integral of |f(x)|, which typically means splitting the interval where f changes sign and taking absolute values.
Piecewise integration
When f(x) is defined by different formulas on different sub-intervals, split the integral at the breakpoints and integrate each piece with its own formula.
Accumulation function from a graph
When F(x) = integral from a to x of f(t) dt and you are given the graph of f, the values of F are computed as signed areas of geometric regions (triangles, rectangles, semicircles, trapezoids) read from the graph.
Find the antiderivative, apply the Evaluation Theorem.
integral from 0 to 1 of (x² + sqrt(x)) dx = [x³/3 + (2/3)x^(3/2)] from 0 to 1 = 1/3 + 2/3 = 1.
integral from pi/6 to 5pi/6 of csc²(theta) d(theta) = [-cot(theta)] from pi/6 to 5pi/6 = -cot(5pi/6) - (-cot(pi/6)) = sqrt(3) + sqrt(3) = 2sqrt(3). (Note: be careful, cot(5pi/6) = -sqrt(3) and cot(pi/6) = sqrt(3).)
integral from 0 to 1 of 4/(t² + 1) dt = [4 arctan(t)] from 0 to 1 = 4(pi/4) - 0 = pi.
For exponential integrands like 3^x, recall the antiderivative of a^x is a^x / ln(a).
(1 - sqrt(u)) / sqrt(u): rewrite as u^(-1/2) - 1, then integrate term by term.
x(2 + x⁵): expand to 2x + x⁶, then integrate.
Always simplify the integrand before hunting for an antiderivative. Splitting fractions, expanding products, and rewriting radicals as powers save time and reduce errors.
integral from 0 to 2 of f(x) dx where f(x) = x⁴ for 0 ≤ x < 1 and f(x) = x⁵ for 1 ≤ x ≤ 2.
Split: integral from 0 to 1 of x⁴ dx + integral from 1 to 2 of x⁵ dx = [x⁵/5] from 0 to 1 + [x⁶/6] from 1 to 2 = 1/5 + (64/6 - 1/6) = 1/5 + 63/6 = 1/5 + 21/2 = 107/10.
For y = x³ - 4x on [-2, 2], the function changes sign. Find the zeros: x(x² - 4) = 0, so x = -2, 0, 2.
On [-2, 0], the function is positive (check a test point). On [0, 2], the function is negative.
Area = integral from -2 to 0 of (x³ - 4x) dx + |integral from 0 to 2 of (x³ - 4x) dx|.
Each piece evaluates to 4, so total area = 4 + 4 = 8.
For y = sqrt(1 - x⁴), the curve is non-negative and meets the x-axis at x = -1 and x = 1. The enclosed area is integral from -1 to 1 of sqrt(1 - x⁴) dx (calculator required).
Given f''(x) = 6x + 12, integrate once: f'(x) = 3x² + 12x + C₁.
Use a tangent-line condition. If f is tangent to 4x - y = 5 at (0, -5), the line y = 4x - 5 has slope 4 and passes through (0, -5). So f'(0) = 4 and f(0) = -5.
f'(0) = 4 gives C₁ = 4. So f'(x) = 3x² + 12x + 4.
Integrate again: f(x) = x³ + 6x² + 4x + C₂.
f(0) = -5 gives C₂ = -5. So f(x) = x³ + 6x² + 4x - 5.
Average value on [-1, 1]: (1/2) integral from -1 to 1 of f(x) dx. Odd-powered terms integrate to zero on symmetric intervals, leaving (1/2) integral from -1 to 1 of (6x² - 5) dx = (1/2)[2x³ - 5x] from -1 to 1 = (1/2)(2 - 5 - (-2 + 5)) = (1/2)(-3 - 3) = -3.
If f(1) = 12, f' is continuous, and integral from 1 to 4 of f'(x) dx = 17, then f(4) = f(1) + 17 = 29.
This is the net change theorem applied directly.
Given dy/dx = sin³(x) and y = 4 when x = 5, find y(7).
y(7) = y(5) + integral from 5 to 7 of sin³(x) dx = 4 + integral from 5 to 7 of sin³(x) dx (use calculator).
For y(0): y(0) = y(5) + integral from 5 to 0 of sin³(x) dx = 4 - integral from 0 to 5 of sin³(x) dx.
General form: y(x) = 4 + integral from 5 to x of sin³(t) dt.
Given the graph of f and F(x) = integral from 2 to x of f(t) dt, compute F at various x-values by calculating signed areas.
F(2) = 0 (integral from 2 to 2).
F(0) = integral from 2 to 0 of f(t) dt = -integral from 0 to 2 of f(t) dt (flip limits, negate).
F has a maximum where f changes from positive to negative (F' = f changes sign from + to -).
F has a minimum where f changes from negative to positive.
F is concave up where f is increasing (F'' = f' > 0), concave down where f is decreasing.
The average value of f on [a, b] is F(b)/(b - a) when F(x) = integral from a to x of f(t) dt.
h'(x) = f(x) by FTC1. Since f has a derivative (negative everywhere), f is differentiable, so h' = f is differentiable, meaning h is twice differentiable. True.
h and dh/dx are both continuous: h is an integral (always continuous), and dh/dx = f is differentiable (hence continuous). True.
Horizontal tangent at x = 1: h'(1) = f(1) = 0. True.
Local maximum at x = 1: h'(1) = 0, and since f' < 0 everywhere, f is strictly decreasing. f passes through zero at x = 1, going from positive (for x < 1) to negative (for x > 1). So h' changes from + to -, giving a local maximum. True.
Local minimum at x = 1: False. It is a local maximum (see above).
Inflection point of h at x = 1: h''(x) = f'(x) < 0 everywhere. Since h'' does not change sign, there is no inflection point. False.
dh/dx crosses the x-axis at x = 1: dh/dx = f(x), and f(1) = 0. Since f is strictly decreasing and passes through zero, it crosses (not just touches) the axis. True.
Signed area = integral from a to b of f(x) dx (can be negative)
Geometric area = integral from a to b of |f(x)| dx (always non-negative)
Piecewise split: integral from a to b = integral from a to c + integral from c to b
Net change recovery: f(b) = f(a) + integral from a to b of f'(x) dx
Average temperature (Newton's Law of Cooling): T_avg = (1/(b - a)) * integral from a to b of T(t) dt
Newton's Law of Cooling provides a clean application: given T(t) = 20 + 75e^(-t/50), the average temperature of coffee over the first 30 minutes is (1/30) * integral from 0 to 30 of (20 + 75e^(-t/50)) dt. The setup is the average value formula applied to a real temperature function, and the answer carries units (°C).
When computing area bounded by a curve and the x-axis, students often just compute the integral from a to b without checking whether the function goes negative. If it does, the integral underestimates the geometric area because negative regions cancel positive ones. Split at the zeros and take absolute values.
Students confuse "F has a maximum" with "f has a maximum." F' = f, so F has a critical point where f = 0, and F has a maximum where f changes from positive to negative. F's extrema come from f's sign changes, not from f's extrema.
On piecewise integrals, students sometimes use the wrong formula on a sub-interval. Always check which piece applies on which part of the domain.
When flipping limits to compute F(0) from F(x) = integral from 2 to x of f(t) dt, students forget the sign change. integral from 2 to 0 = -integral from 0 to 2.
⚠️ "Evaluate without a calculator" problems test antiderivative fluency. Know your trig antiderivatives (especially csc², sec tan, and inverse trig forms like 1/(1 + t²)) cold.
⚠️ Area problems where the curve crosses the x-axis are a classic trap. The integral alone does not give the area unless the function is non-negative on the whole interval.
⚠️ Graph-based accumulation problems (given graph of f, analyse F = integral of f) are a staple of AP free-response sections. You need to connect f's zeros to F's extrema, and f's increasing/decreasing behaviour to F's concavity.
⚠️ The Newton's Law of Cooling average-temperature problem is a textbook application that combines average value with exponential functions. Show the integral setup and include units.
Fill in the blank: integral from 0 to 1 of 4/(t² + 1) dt = ______.
pi.
True or false: If f(x) is negative on [a, b], then the integral from a to b of f(x) dx gives the area between f and the x-axis.
False. The integral gives a negative number. The area is the absolute value of the integral.
Fill in the blank: If f(1) = 12 and integral from 1 to 4 of f'(x) dx = 17, then f(4) = ______.
True or false: If F(x) = integral from a to x of f(t) dt, then F has a local maximum wherever f has a local maximum.
False. F has a local maximum where f changes from positive to negative (where f = 0 and f is decreasing through zero), not where f has its own peak.
Q: (Calculator) What is the area enclosed between y = sqrt(1 - x⁴) and the x-axis?
A: The curve is non-negative, meeting the x-axis at x = -1 and x = 1. Area = integral from -1 to 1 of sqrt(1 - x⁴) dx ≈ 1.748. Answer: (E) 1.748.
Q: Find the area bounded by the x-axis and y = x³ - 4x on [-2, 2].
A: Zeros at x = -2, 0, 2. Integral from -2 to 0 of (x³ - 4x) dx = [x⁴/4 - 2x²] from -2 to 0 = 0 - (4 - 8) = 4. Integral from 0 to 2 of (x³ - 4x) dx = (4 - 8) - 0 = -4. Area = 4 + |-4| = 8.
Q: Evaluate integral from 0 to 1 of (x² + sqrt(x)) dx.
A: [x³/3 + (2/3)x^(3/2)] from 0 to 1 = 1/3 + 2/3 = 1.
Q: If f(1) = 12, f' is continuous, and integral from 1 to 4 of f'(x) dx = 17, what is f(4)?
A: f(4) = f(1) + 17 = 12 + 17 = 29.
Q: Given dy/dx = sin³(x) and y(5) = 4, write an expression for y(x).
A: y(x) = 4 + integral from 5 to x of sin³(t) dt.
Q: (Calculator) Average temperature of coffee with T(t) = 20 + 75e^(-t/50) over the first 30 minutes?
A: T_avg = (1/30) * integral from 0 to 30 of (20 + 75e^(-t/50)) dt. The antiderivative is 20t - 3750e^(-t/50). Evaluating: [600 - 3750e^(-0.6)] - [0 - 3750] = 600 - 3750(0.5488) + 3750 = 600 - 2058.0 + 3750 = 2292.0. Divide by 30: ≈ 76.4°C.
Evaluating definite integrals feeds directly into applications chapters: area between curves, volumes by disc/washer/shell methods, and work/force integrals all start with setting up and evaluating a definite integral. The graph-based accumulation analysis connects to the curve-sketching techniques from the derivatives unit (sign charts, concavity tests), applied now to the relationship F' = f rather than to f' and f. Newton's Law of Cooling ties into the separable differential equations topic.
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