Difficulty: Intermediate | Prerequisites: Basic differentiation rules, antiderivative concept
The Fundamental Theorem of Calculus (FTC) connects differentiation and integration. Part 1 says the derivative of an accumulation function equals the integrand evaluated at the upper limit. Part 2 says you can evaluate a definite integral by finding any antiderivative and computing the difference at the endpoints. These two ideas let you move freely between derivatives and integrals, which is the backbone of integral calculus.
Definite integral
The integral of a function f from a to b, written as the integral from a to b of f(x) dx. It gives the net signed area between the graph of f and the x-axis on [a, b].
In simple terms, it is the total accumulated quantity of f over an interval, where area below the x-axis counts as negative.
Antiderivative (indefinite integral)
A function F whose derivative equals f. That is, F'(x) = f(x). The most general antiderivative includes an arbitrary constant C.
Think of it as "undoing" a derivative: if differentiating F gives you f, then F is an antiderivative of f.
Fundamental Theorem of Calculus, Part 1 (FTC1)
If f is continuous on [a, b] and F(x) = the integral from a to x of f(t) dt, then F'(x) = f(x).
In simple terms, differentiating an accumulation function just gives you back the original function.
Fundamental Theorem of Calculus, Part 2 (FTC2)
If f is continuous on [a, b] and F is any antiderivative of f, then the integral from a to b of f(x) dx = F(b) - F(a).
In simple terms, to evaluate a definite integral, find any antiderivative, plug in the top and bottom limits, and subtract.
Accumulation function
A function defined as f(x) = the integral from a to x of h(t) dt. Its value at x equals the net signed area under h from a to x.
Think of it as a running total: as x moves right, the function accumulates (or loses) area depending on whether h is positive or negative.
Critical points of an accumulation function
Since the derivative of the accumulation function equals the integrand (by FTC1), the critical points occur where the integrand equals zero, that is, at the x-intercepts of the integrand's graph.
Graphical interpretation of accumulation functions
If h(t) is given as a graph and f(x) = the integral from a to x of h(t) dt, then f(x) represents the net signed area under h from a to x.
f(a) = 0 always, because the integral from a to a is zero. This is a quick way to eliminate wrong graph choices.
f is only defined for x at or above the lower limit. If the lower limit is -2, then there is no graph of f to the left of x = -2.
Increasing and decreasing behaviour
By FTC1, f'(x) = h(x). So where h is positive, f is increasing. Where h is negative, f is decreasing.
To identify the correct graph of f from multiple choices, check whether f's slope matches the sign of h on each interval.
Finding maxima and minima of accumulation functions
If g(x) = the integral from a to x of f(t) dt, then g'(x) = f(x).
Critical points of g occur at the zeros of f, that is, at the x-intercepts of f's graph.
Use a sign chart for g' = f to determine which critical points are maxima (g' goes from + to -) and which are minima (g' goes from - to +).
FTC1 with chain rule (composite upper limits)
When the upper limit is not simply x but a function of x, such as x squared, combine FTC1 with the chain rule.
Example: if F(x) = d/dx of the integral from 0 to x squared of 5t^6 dt, first evaluate the integral to get (5/7)x^14, then differentiate to get F(x) = 10x^13, so F(1) = 10.
Second derivatives via FTC1
If F(x) = the integral from 0 to x of some function, then F'(x) equals that function. To find F''(x), differentiate F'(x) using standard rules (chain rule, product rule, etc.).
Example: F(x) = the integral from 0 to x of 4e^(-6cos^2(theta)) d(theta). Then F'(x) = 4e^(-6cos^2(x)), and F''(x) requires the chain rule on the exponential and trigonometric composition.
The basic method
To evaluate the integral from a to b of f(x) dx, find any antiderivative F(x) of f(x), then compute F(b) - F(a).
You do not need the most general antiderivative (the +C cancels out), so pick the simplest one.
Polynomial integrands
Use the power rule for antiderivatives: the antiderivative of x^r is x^(r+1)/(r+1), provided r is not -1.
Example: the integral from 0 to 2 of (1 + 6y - y^2) dy. The antiderivative is y + 3y^2 - (1/3)y^3. Evaluating at 2 and subtracting the value at 0 gives 2 + 12 - 8/3 = 34/3.
Example: the integral from 0 to 4 of (5x - 3 times the square root of x) dx. Write the square root of x as x^(1/2) and use the power rule. The antiderivative is (5/2)x^2 - 2x^(3/2). The result is 40 - 16 = 24.
Fractional exponents and x^(e^2)
The power rule applies to any real exponent except -1. For x^(e^2), the antiderivative is x^(e^2 + 1)/(e^2 + 1).
Example: the integral from 0 to 1 of (x^(e^2) - 4x) dx = 1/(1 + e^2) - 2 = -(1 + 2e^2)/(1 + e^2).
Trigonometric integrands
Key antiderivatives: the antiderivative of cos(x) is sin(x); the antiderivative of sin(x) is -cos(x); the antiderivative of sec^2(x) is tan(x).
Example: the integral from 0 to pi/2 of (4cos(x) - 3sin(x)) dx. The antiderivative is 4sin(x) + 3cos(x). Evaluating gives (4 times 1 + 3 times 0) - (4 times 0 + 3 times 1) = 1.
Example: the integral from 0 to pi/6 of (3sec^2(theta) + 2cos(theta)) d(theta). The antiderivative is 3tan(theta) + 2sin(theta). The result is 3/sqrt(3) + 1 = sqrt(3) + 1.
Exponential integrands
The antiderivative of e^(ax) is e^(ax)/a.
Example: the integral from 0 to ln(2) of e^(3y) dy = (1/3)e^(3y) evaluated from 0 to ln(2) = (1/3)(e^(3ln2) - 1) = (1/3)(8 - 1) = 7/3.
Useful identity: e^(k ln(a)) = a^k. This comes up frequently when the upper limit involves a logarithm.
Logarithmic integrands
The antiderivative of 1/x is ln|x|.
Example: the integral from 1 to 2 of (x + 4/x) dx = (1/2)x^2 + 4ln(x) evaluated from 1 to 2 = 3/2 + 4ln(2).
Constant multiple rule
The integral from a to b of m times g(x) dx = m times the integral from a to b of g(x) dx, for any constant m.
If G is an antiderivative of g (so G'(x) = g(x)), then mG is an antiderivative of mg, and the result is m(G(b) - G(a)).
A common mistake is treating m as though it depends on x. The constant multiple rule only applies when m is truly constant.
Area under a curve
"Area under the graph" on [a, b] means the integral from a to b of f(x) dx, provided f(x) is non-negative on that interval.
If f(x) takes negative values, the integral gives net signed area, not geometric area. For geometric area, integrate the absolute value of f.
Example: the area under f(x) = (x^2 + 4)/x on [1, 2]. Rewrite as x + 4/x, then integrate to get (1/2)x^2 + 4ln(x). The result is 3/2 + 4ln(2).
\int x^r \, dx = \frac{x^{r+1}}{r+1} + C \quad (r \neq -1)\int \frac{1}{x} \, dx = \ln|x| + C\int e^{ax} \, dx = \frac{e^{ax}}{a} + C\int \cos x \, dx = \sin x + C\int \sin x \, dx = -\cos x + C\int \sec^2 x \, dx = \tan x + C\frac{d}{dx} \int_a^x f(t) \, dt = f(x) \quad \text{(FTC Part 1)}\int_a^b f(x) \, dx = F(b) - F(a) \quad \text{where } F'(x) = f(x) \quad \text{(FTC Part 2)}Useful identity: e^(k ln a) = a^k. This converts expressions like e^(3 ln 2) into 2^3 = 8.
Students often forget that f(a) = 0 when f is defined as the integral from a to x. Any proposed graph of f that does not pass through zero at x = a is wrong.
Students sometimes apply FTC1 without the chain rule when the upper limit is a function of x (such as x squared). If the upper limit is u(x), the derivative picks up an extra factor of u'(x).
A common error is forgetting that the integral gives net signed area, not absolute area. If f dips below the x-axis, those portions subtract from the total.
When evaluating F(b) - F(0), students sometimes forget that terms like ln(1) = 0 or e^0 = 1, leading to arithmetic mistakes in the final answer.
Expect questions that give you a graph of h and ask you to identify the graph of f(x) = the integral from a to x of h(t) dt. Eliminate choices by checking: f(a) = 0, f only defined for x at or above a, and f increasing where h is positive.
Finding maxima and minima of accumulation functions is heavily tested. You need the sign chart of g' = f to determine which zeros of f give maxima of g.
Evaluating definite integrals with polynomial, trig, exponential, and logarithmic integrands is standard exam material. Be fluent with all the basic antiderivative formulas.
FTC1 combined with the chain rule (composite upper limits) appears regularly. Remember to multiply by the derivative of the upper limit.
The constant multiple rule for integrals is often tested as a conceptual question, not just a computation.
True or false: If f(x) = the integral from 3 to x of g(t) dt, then f(3) = 1. Answer: False. f(3) = 0 always.
Fill in the blank: By FTC1, if F(x) = the integral from 0 to x of e^(t^2) dt, then F'(x) = ______. Answer: e^(x^2).
True or false: The integral from a to b of 5 times f(x) dx = 5 times the integral from a to b of f(x) dx. Answer: True.
Fill in the blank: The antiderivative of sec^2(x) is ______. Answer: tan(x) + C.
True or false: e^(2 ln 3) = 6. Answer: False. e^(2 ln 3) = 3^2 = 9.
Q: If g(x) = the integral from 2 to x of f(t) dt, and f changes sign from positive to negative at x = 5, what can you say about g at x = 5?
A: g has a local maximum at x = 5, because g'(x) = f(x) changes from positive to negative there.
Q: Evaluate the integral from 0 to ln(2) of e^(3y) dy.
A: (1/3)(e^(3 ln 2) - 1) = (1/3)(8 - 1) = 7/3.
Q: If F(x) = d/dx of the integral from 0 to x^2 of 5t^6 dt, what is F(1)?
A: The integral equals (5/7)x^14. Differentiating gives F(x) = 10x^13, so F(1) = 10.
Q: Evaluate the integral from 0 to pi/2 of (4cos(x) - 3sin(x)) dx.
A: The antiderivative is 4sin(x) + 3cos(x). At pi/2: 4(1) + 3(0) = 4. At 0: 4(0) + 3(1) = 3. The result is 4 - 3 = 1.
Q: If g(x) is continuous on [a, b] and m is a constant, express the integral from a to b of m times g(x) dx in terms of an antiderivative G of g.
A: m(G(b) - G(a)), where G'(x) = g(x).
This material connects directly to differentiation rules (chain rule, product rule, quotient rule), since FTC1 turns integration questions into differentiation questions. Understanding antiderivatives here prepares you for u-substitution and integration by parts, which extend the range of functions you can integrate.
The idea of accumulation functions also appears in applied contexts such as net change problems (distance from velocity, total cost from marginal cost), which are standard applications in later chapters.
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