Step 21: The Fundamental Theorem of Calculus
- An antiderivative of \( f \) is a function \( F \) with \( F’ = f \). They come in families: \( \int f(x)\,dx = F(x) + C \). One known point fixes \( C \).
- FTC Part 2 turns a limit of sums into one subtraction: \( \int_a^b f(x)\,dx = F(b)-F(a) \). Rates integrate to total change, so marginal cost integrates to extra cost.
- FTC Part 1 says the accumulation function \( g(x) = \int_a^x f(t)\,dt \) has derivative \( f(x) \). With an upper limit \( u(x) \), multiply by \( u'(x) \).
- Exam point: on AP graph questions, \( g’ = f \). So \( g \) rises where \( f \) is above the axis, peaks where \( f \) crosses from positive to negative, and is concave up where \( f \) is increasing. Name the function in every sentence.
Antiderivatives and the constant of integration
Step 20 found the exact area under \( y = 9-x^2 \) from 0 to 3 by adding infinitely many rectangles, with a page of algebra. This step shows that the same answer, 18, comes from two lines of working. The shortcut needs one new skill: given \( f \), find a function whose derivative is \( f \).
Why the \( + C \)? If \( F’ = f \), then \( F + 5 \) and \( F-12 \) have the same derivative, because a constant differentiates to zero. On an interval these are the only possibilities: two functions with the same derivative differ by a constant, a consequence of the Mean Value Theorem from Step 17.
Every rule is a derivative read backwards.
| \( f(x) \) | \( \int f(x)\,dx \) | Comes from |
|---|---|---|
| \( x^n \), \( n \neq -1 \) | \( \dfrac{x^{n+1}}{n+1} + C \) | raise the power by one, divide by the new power |
| \( \dfrac1x \) | \( \ln|x| + C \) | fills the gap at \( n = -1 \) |
| \( e^x \) | \( e^x + C \) | \( e^x \) is its own derivative |
| \( \cos x \) | \( \sin x + C \) | \( \frac{d}{dx}\sin x = \cos x \) |
| \( \sin x \) | \( -\cos x + C \) | \( \frac{d}{dx}\cos x = -\sin x \) |
| \( \sec^2 x \) | \( \tan x + C \) | \( \frac{d}{dx}\tan x = \sec^2 x \) |
Constants and sums pass straight through: \( \int [k f(x) \pm g(x)]\,dx = k\int f(x)\,dx \pm \int g(x)\,dx \). Products and quotients do not, so rewrite them first.
Find (a) \( \int \left(8x^3-\dfrac{3}{x^4} + 6\sqrt{x}\right)dx \), (b) \( \int (x + 3)(2x-1)\,dx \) and (c) \( \int (2\cos x + 5\sin x-e^x)\,dx \).
Differentiating your answer in your head is the quickest check there is, and it catches almost every sign slip.
An indefinite integral is a family of curves, each a vertical shift of the others. A single known point picks out one member. If \( \dfrac{dy}{dx} = 6x^2-2x + 5 \) and the curve passes through \( (1, 4) \), then \( y = 2x^3-x^2 + 5x + c \), and \( 4 = 2-1 + 5 + c \) gives \( c = -2 \). The curve is \( y = 2x^3-x^2 + 5x-2 \).
FTC Part 2: evaluating definite integrals
Here is the theorem that joins the two halves of calculus.
Any antiderivative will do, because the constant cancels: \( (F(b) + C)-(F(a) + C) = F(b)-F(a) \). So in definite integrals you can leave \( C \) out.
The theorem has a natural reading. If \( F’ = f \), then \( f \) is the rate of change of \( F \), and Step 20 showed that integrating a rate gives the total change. So \( \int_a^b F'(x)\,dx = F(b)-F(a) \): the accumulated change in \( F \) from \( a \) to \( b \). AP calls this the net change theorem, and it is how free-response questions want you to use a rate: “amount at the end = amount at the start + integral of the rate”.
Evaluate (a) \( \int_1^2 (6x^2-4x + 1)\,dx \), (b) \( \int_0^{\pi/3} \sin x\,dx \), (c) \( \int_1^4 \dfrac{1}{\sqrt{x}}\,dx \) and (d) \( \int_0^3 (9-x^2)\,dx \).
Part (d) is the area that took a limit of Riemann sums in Step 20. Now it takes one line.
An economics example: total cost from marginal cost. Marginal cost is the derivative of total cost, so total cost is an antiderivative of marginal cost, and the constant is the fixed cost (the cost of producing nothing). Suppose \( MC = 12 + 0.6Q \) dollars per unit and fixed cost is $300. Then \( TC = 300 + 12Q + 0.3Q^2 \), and 40 units cost \( 300 + 480 + 480 = \$1260 \). FTC Part 2 answers a sharper question directly: the extra cost of raising output from 20 to 40 units is
The fixed cost never appears, because it cancels in the subtraction, which is exactly right: it is paid whatever the output.
FTC Part 1: differentiating accumulation functions
Fix a starting point \( a \) and let the right-hand edge of the region move. The signed area from \( a \) up to \( x \) is a function of \( x \):
The letter \( t \) is only a placeholder inside the integral; \( x \) is the moving edge. This is an accumulation function. How fast does it grow? Push the edge from \( x \) to \( x + h \). The extra area is a thin strip of width \( h \) and height very nearly \( f(x) \) (Figure 1), so \( \dfrac{g(x+h)-g(x)}{h} \approx f(x) \), and the approximation becomes exact as \( h \to 0 \).
When the upper limit is a function \( u(x) \), the chain rule adds a factor, and swapping the limits adds a minus sign:
Find (a) \( \dfrac{d}{dx}\displaystyle\int_1^x \cos(t^3)\,dt \), (b) \( \dfrac{d}{dx}\displaystyle\int_0^{x^3} \sqrt{1 + t^2}\,dt \) and (c) \( \dfrac{d}{dx}\displaystyle\int_x^4 e^{t^2}\,dt \).
The lower limit 1, 0 or 4 never matters to the derivative: changing it only adds a constant to \( g \).
Stocks and flows. Economics is full of accumulation functions. Net investment \( I(t) \) is the rate at which a firm’s capital stock grows. If capital starts at $200 million and investment runs at \( I(t) = 30\sqrt{t} \) million dollars per year, the capital stock after \( t \) years is \( K(t) = 200 + \int_0^t 30\sqrt{s}\,ds = 200 + 20t^{3/2} \). After 9 years, \( K(9) = 200 + 20 \times 27 = \$740 \) million. And Part 1 gives \( K'(t) = 30\sqrt{t} = I(t) \): a stock grows at exactly the rate of its flow.
Reading an accumulation function from a graph
AP free-response questions love this set-up: you are given the graph of \( f \) and asked about \( g(x) = \int_a^x f(t)\,dt \). Since \( g’ = f \) and \( g” = f’ \), everything about the shape of \( g \) can be read off the graph of \( f \), and the areas give its values.
| You want, about \( g \) | Look at, on the graph of \( f \) |
|---|---|
| values of \( g \) | signed area from \( a \) to \( x \); \( g(a) = 0 \) |
| where \( g \) increases | where \( f > 0 \) (above the axis) |
| relative maximum of \( g \) | where \( f \) changes from positive to negative |
| where \( g \) is concave up | where \( f \) is increasing |
| inflection point of \( g \) | where \( f \) has a turning point |
Let \( f(t) = t^2-6t + 8 \) and \( g(x) = \int_0^x f(t)\,dt \) for \( 0 \le x \le 6 \). (a) Where is \( g \) increasing? (b) Find and classify the relative extrema of \( g \). (c) Where is \( g \) concave up? (d) Find the absolute maximum and minimum values of \( g \).
The absolute maximum is 12, at \( x = 6 \), and the absolute minimum is 0, at \( x = 0 \). The relative minimum at \( x = 4 \) stays positive because the area lost on \( [2, 4] \), which is \( \tfrac43 \), is smaller than the \( \tfrac{20}{3} \) gained on \( [0, 2] \).
Functions of ax + b, and simple improper integrals
What is \( \int (4x + 1)^5\,dx \)? Differentiating \( (4x + 1)^6 \) gives \( 6(4x + 1)^5 \times 4 = 24(4x + 1)^5 \), which is 24 times too big. So \( \int (4x + 1)^5\,dx = \tfrac{1}{24}(4x + 1)^6 + C \). The general rule: if \( \int f(x)\,dx = F(x) + C \), then
So \( \int e^{ax+b}\,dx = \tfrac1a e^{ax+b} + C \), \( \int \dfrac{1}{ax + b}\,dx = \tfrac1a \ln|ax + b| + C \), and \( \int \cos(ax + b)\,dx = \tfrac1a\sin(ax + b) + C \). This works only because the inner derivative \( a \) is a constant. For \( (x^2 + 1)^3 \) the inner derivative is \( 2x \), and you need substitution (Step 22).
Find (a) \( \int \dfrac{6}{3x-2}\,dx \) and (b) \( \int \sin(4-x)\,dx \), and evaluate (c) \( \int_0^{\pi/4} \cos 2x\,dx \).
In (b) the two minus signs cancel. Checking by differentiation confirms it: \( \frac{d}{dx}\cos(4-x) = -\sin(4-x) \times (-1) = \sin(4-x) \).
Finally, can a region that runs off to infinity have a finite area? Sometimes. An improper integral has an infinite limit, or an integrand that blows up at an end of the interval, and it is defined as a limit of ordinary integrals. Improper integrals are AP BC only (topic 6.13); Cambridge Paper 1 includes simple cases like these. For example,
The integrand \( x^{-1/3} \) is unbounded near 0, yet \( \int_0^8 x^{-1/3}\,dx = \lim_{t \to 0^+}\Big[\tfrac32x^{2/3}\Big]_t^8 = 6-0 = 6 \). Not every such integral converges: \( \int_1^b x^{-1/2}\,dx = 2\sqrt{b}-2 \) grows without limit, so \( \int_1^\infty x^{-1/2}\,dx \) diverges. Step 23 treats improper integrals in full.
Practise: 6 questions
Try each one on paper first, then check the worked answer.
EasyQ1. Find \( \int \left(10x^4-\dfrac{2}{x^3} + 1\right)dx \). Show worked answer
Integrate: \( \dfrac{10x^5}{5}-\dfrac{2x^{-2}}{-2} + x + C \).
\( = 2x^5 + \dfrac{1}{x^2} + x + C \).
EasyQ2. Evaluate \( \int_1^3 (x^2 + 2x)\,dx \). Show worked answer
\( = 18-\tfrac43 = \tfrac{50}{3} \).
MediumQ3. A curve has \( \dfrac{dy}{dx} = 3\sqrt{x}-2 \) and passes through \( (4, 10) \). Find its equation. Show worked answer
At \( (4, 10) \): \( 10 = 2(8)-8 + c \), so \( c = 2 \).
The curve is \( y = 2x^{3/2}-2x + 2 \).
MediumQ4. Find \( \dfrac{d}{dx}\displaystyle\int_2^{x^2 + 1} \ln t\,dt \). Show worked answer
Derivative \( = \ln(u)\cdot u’ = 2x\ln(x^2 + 1) \).
MediumQ5. Evaluate (a) \( \int_0^1 (2x + 1)^3\,dx \) and (b) \( \int_0^{\ln 2} e^{2x}\,dx \). Show worked answer
(b) \( \Big[\tfrac12e^{2x}\Big]_0^{\ln 2} = \tfrac12\left(e^{2\ln 2}-1\right) = \tfrac12(4-1) = \tfrac32 \).
ExamQ6. Let \( f(t) = t^2-6t + 5 \) and \( g(x) = \int_0^x f(t)\,dt \) for \( 0 \le x \le 6 \). (a) Find \( g'(3) \) and \( g”(3) \). [2] (b) Find the \( x \)-coordinates of the relative maximum and relative minimum of \( g \), justifying your answers. [2] (c) Find the absolute maximum value of \( g \) on \( [0, 6] \). Justify your answer. [3] (d) Let \( h(x) = \int_0^{2x} f(t)\,dt \). Find \( h'(1) \). [2] Show worked answer
\( g”(3) = f'(3) = 2(3)-6 = 0 \). [B1]
(b) \( f(t) = (t-1)(t-5) \). \( g’ = f \) changes from positive to negative at \( x = 1 \): relative maximum. [B1]
\( g’ = f \) changes from negative to positive at \( x = 5 \): relative minimum. [B1]
(c) \( g(x) = \tfrac13x^3-3x^2 + 5x \). Candidates: endpoints and the critical points. [M1]
\( g(0) = 0 \), \( g(1) = \tfrac13-3 + 5 = \tfrac73 \), \( g(5) = \tfrac{125}{3}-75 + 25 = -\tfrac{25}{3} \), \( g(6) = 72-108 + 30 = -6 \). [A1]
Absolute maximum value \( \tfrac73 \), at \( x = 1 \). [A1]
(d) \( h'(x) = f(2x)\cdot 2 \), so \( h'(1) = 2f(2) = 2(4-12 + 5) = -6 \). [M1 A1]
Total: 9 marks.
Check yourself
1. What is \( \dfrac{d}{dx}\displaystyle\int_3^x \sqrt{t^2 + 1}\,dt \)?
2. What is \( \int_0^2 (3x^2 + 1)\,dx \)?
3. Which is \( \int e^{5x}\,dx \)?
Your score: 0 out of 3
The short version
- \( \int x^n\,dx = \tfrac{x^{n+1}}{n+1} + C \) (\( n \neq -1 \)), \( \int \tfrac1x\,dx = \ln|x| + C \); a known point fixes \( C \). Expand products first.
- FTC Part 2: \( \int_a^b f(x)\,dx = F(b)-F(a) \) for any antiderivative \( F \). Integrating a rate gives the net change, so \( TC = FC + \int MC\,dQ \).
- FTC Part 1: \( \frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f(u(x))\,u'(x) \); with \( x \) as the lower limit, add a minus sign.
- For \( g(x) = \int_a^x f(t)\,dt \): \( g \) rises where \( f > 0 \), and is concave up where \( f \) rises. Check endpoints for absolute extrema.
- \( \int f(ax + b)\,dx = \tfrac1aF(ax + b) + C \). Improper integrals are limits of ordinary ones (AP BC; Cambridge P1 simple cases).
- The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.
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