Difficulty: Intermediate to Advanced | Prerequisites: FTC Part I differentiation (see companion notes), concavity and inflection points, reading graphs of functions
This picks up where the basic FTC I differentiation notes leave off. Once you can find g'(x) from an integral-defined function, the next layer is analysing what g looks like: where is it increasing, where is it concave up, where are its inflection points? These questions dominate AP Calculus free-response sections. You need to be comfortable reading a graph of f and translating that into information about g, g', and g''.
When g is defined as an integral of f, you can read everything about g from f: g' = f tells you where g increases, and g'' = f' tells you where g is concave up or down. AP free-response problems hand you a graph of f and ask you to extract g(a), g'(a), g''(a), average rates of change, and inflection points, all without ever computing the integral.
Concave up
A function is concave up on an interval where its second derivative is positive: g''(x) > 0. Think of it as the curve opening upwards, like a bowl. The slope is increasing.
Concave down
A function is concave down on an interval where its second derivative is negative: g''(x) < 0. The curve opens downwards, like an upside-down bowl. The slope is decreasing.
Inflection point
A point where the concavity changes, from up to down or down to up. At an inflection point, g''(x) = 0 or is undefined, and g'' changes sign. In simple terms, it is where the curve switches from bending one way to bending the other.
Average rate of change
The slope of the secant line between two points: [g(b) – g(a)] / (b – a). Think of it as the "overall slope" between two x-values, ignoring what happens in between.
Mean Value Theorem (MVT) connection
If g is continuous on [a, b] and differentiable on (a, b), there exists at least one c in (a, b) where g'(c) equals the average rate of change. This is how you justify that g' hits a particular value somewhere in the interval.
The key chain of reasoning:
g(x) = ∫ from a to x of f(t) dt
g'(x) = f(x) (by FTC I)
g''(x) = f'(x) (differentiate again)
So the concavity of g depends on the sign of f'(x), the derivative of the original integrand.
g is concave up where f'(x) > 0, meaning where f is increasing
g is concave down where f'(x) < 0, meaning where f is decreasing
g has an inflection point where f' changes sign, meaning where f has a local maximum or minimum
From the worksheet: g(x) = ∫ from 0 to x of 1/(1 + t + t²) dt. Find where g is concave up.
Step 1: g'(x) = 1/(1 + x + x²) by FTC I.
Step 2: Find g''(x). Rewrite g'(x) = (1 + x + x²)⁻¹ and differentiate:
g''(x) = –(1 + x + x²)⁻² · (1 + 2x)
Step 3: Set g''(x) = 0. The numerator –(1 + 2x) = 0 gives x = –1/2.
Step 4: Sign analysis. g''(x) > 0 when –(1 + 2x) > 0, which is when 1 + 2x < 0, so x < –1/2.
Conclusion: g is concave up on (–∞, –1/2).
Given: f is defined on [0, 7], consisting of four line segments (shown in graph). g(x) = ∫ from 2 to x of f(t) dt.
Part (a): Find g(3), g'(3), and g''(3).
g(3) = ∫ from 2 to 3 of f(t) dt = area of a triangle with base 1 and height (4 + 2)/2. Reading the graph: the region from t = 2 to t = 3 forms a triangle with area = (1/2)(4 + 2)(1) = 3. So g(3) = 3.
g'(3) = f(3) = 2 (read directly from the graph of f at x = 3).
g''(3) = f'(3) = slope of f at x = 3. From the graph, f is a line segment with slope = –2 near x = 3. So g''(3) = –2.
Part (b): Average rate of change of g on [0, 3].
Need g(0) and g(3). g(0) = ∫ from 2 to 0 of f(t) dt = –∫ from 0 to 2 of f(t) dt.
From the graph, ∫ from 0 to 2 of f(t) dt = area of a triangle = (1/2)(2)(4) = 4. So g(0) = –4.
Average rate = [g(3) – g(0)] / (3 – 0) = [3 – (–4)] / 3 = 7/3.
Part (c): How many times does g'(c) = 7/3 for 0 < c < 3?
Since g'(x) = f(x), we need f(c) = 7/3. The line y = 7/3 intersects the graph of f twice on (0, 3). So the answer is twice. This is justified by the fact that f is continuous and crosses the value 7/3 at two points in that interval.
Part (d): Inflection points of g on (0, 7).
g''(x) = f'(x), so inflection points of g occur where f' changes sign, which is where f changes from increasing to decreasing or vice versa.
From the graph, f has local extrema (slope changes sign) at approximately x = 2 and x = 5. So the inflection points of g are at x = 2 and x = 5.
Concavity of an accumulation function
If g(x) = ∫ from a to x of f(t) dt, then g''(x) = f'(x). Concave up where f'(x) > 0. Concave down where f'(x) < 0.
Average rate of change
[g(b) – g(a)] / (b – a)
Evaluating g(a) from a graph
g(a) = ∫ from c to a of f(t) dt = net signed area under f from c to a. Triangles: (1/2)(base)(height). Trapezoids: (1/2)(b1 + b2)(h). Count area below the x-axis as negative.
If f(t) represents the rate at which water flows into a tank, then g(x) = ∫ from a to x of f(t) dt is the total volume of water added up to time x. The concavity of g tells you whether the flow rate is speeding up (concave up) or slowing down (concave down), which matters for flood modelling and reservoir management.
In economics, if f(t) is a company's revenue growth rate, the concavity of the accumulated revenue function tells you whether growth is accelerating or decelerating, a distinction that drives investment decisions.
Students often confuse where f = 0 with where g has an inflection point. f = 0 tells you where g has a critical point (potential max/min), not where g changes concavity. Inflection points of g occur where f' changes sign.
Students read the graph of f and report it as the graph of g. If you are given f, then g' = f, not g = f. The graph you see is the derivative of the function you are analysing.
Students compute the average rate of change using f values instead of g values. The average rate of change of g uses g(b) and g(a), which are areas under f, not the heights of f.
Students say g'' does not exist at corners of f. If f has a corner (is not differentiable), then g'' = f' does not exist there, but that does not automatically make it an inflection point. You need f' to change sign.
⚠️ The "graph of f, define g as an integral" format is one of the most reliable AP Calculus free-response patterns. It has appeared on the AB and BC exams repeatedly (2003B #5, 2008 #5, 2010 #4, among others).
⚠️ Exams nearly always ask for g(a), g'(a), and g''(a) in the same problem. Practise computing all three from a single graph.
⚠️ Justification matters. Saying "g is concave up because g'' > 0" earns more credit than just stating the interval. Always tie your answer back to the sign of g'' = f'.
⚠️ Average rate of change plus MVT is a common follow-up: they ask how many times g' equals the average rate. Use the Intermediate Value Theorem on f (since g' = f) to count crossings.
True or false: If g(x) = ∫ from a to x of f(t) dt, then g is concave up wherever f is positive. (False. Concave up is where f' > 0, i.e. where f is increasing, not where f is positive.)
Fill in the blank: Inflection points of g occur where ____ changes sign. (f', or equivalently where f has a local max or min.)
True or false: g(a) always equals zero when a is the lower limit of integration. (True. ∫ from a to a of f(t) dt = 0.)
Fill in the blank: The average rate of change of g on [a, b] equals [g(b) – g(a)] / ____. (b – a)
True or false: If f has a corner at x = c, then g necessarily has an inflection point at x = c. (False. A corner means f' is undefined there, but you still need f' to change sign around c.)
Q: Let g(x) = ∫ from 0 to x of 1/(1 + t + t²) dt. On what intervals is g concave up?
A: g'(x) = 1/(1 + x + x²). Then g''(x) = –(1 + 2x)/(1 + x + x²)². Setting the numerator equal to zero: 1 + 2x = 0, so x = –1/2. Since g''(x) > 0 when x < –1/2, g is concave up on (–∞, –1/2).
Q: Given the graph of f on [0, 7] and g(x) = ∫ from 2 to x of f(t) dt, find g(3).
A: g(3) = ∫ from 2 to 3 of f(t) dt. Read the graph: the area from t = 2 to t = 3 is a trapezoid (or triangle, depending on the graph). Compute the signed area using geometric formulas. In the 2003B problem, g(3) = 3.
Q: Using the same setup, find the average rate of change of g on [0, 3].
A: Compute g(0) = ∫ from 2 to 0 of f(t) dt = –∫ from 0 to 2 of f(t) dt = –(area under f from 0 to 2). Then average rate = [g(3) – g(0)] / 3. In the 2003B problem, g(0) = –4, so the average rate = (3 – (–4))/3 = 7/3.
Q: How do you find inflection points of g when given a graph of f?
A: Since g''(x) = f'(x), inflection points of g are where f' changes sign. Look at the graph of f for local maxima and minima; those x-values are the inflection points of g. In the 2003B problem, f changes from increasing to decreasing at x = 2 and from decreasing to increasing at x = 5, so g has inflection points at x = 2 and x = 5.
Q: In the 2003B problem, for how many values c in (0, 3) is g'(c) = 7/3? Justify.
A: Since g'(x) = f(x), we need f(c) = 7/3. Draw the horizontal line y = 7/3 on the graph of f. Count how many times it intersects f on (0, 3). The line crosses twice, so the answer is two. The justification relies on f being continuous and the Intermediate Value Theorem.
This connects directly to curve sketching from earlier in Calculus I: the same concavity and inflection-point analysis you learned for explicit functions applies here, except now the function is defined implicitly through an integral.
The Mean Value Theorem appears whenever an exam asks "for how many values does g' equal the average rate of change." The reasoning is identical to the MVT applications you practised in the derivatives unit.
FTC Part II (evaluating definite integrals using antiderivatives) is the other half of the Fundamental Theorem. Part I goes from integral to derivative; Part II goes from derivative to integral. They are complementary, and exam questions sometimes require both in the same problem.
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