Difficulty: Intermediate to Advanced | Prerequisites: Limits and Continuity notes, Differentiation Rules notes
This material ties differentiation to real questions (where are the peaks and troughs of a function? where does the curvature change?) and introduces integration as the reverse process. Optimisation problems and definite integrals are staples of any Calculus 1 final.
Derivatives tell you where a function is climbing or falling, and where its curvature changes direction. Integration reverses differentiation: it lets you add up infinitely many thin slices to find areas, accumulated quantities, and antiderivatives. Both sides show up on every Calculus 1 exam, and u-substitution is the single most important integration technique at this level.
Critical point
A value of x where f'(x) = 0 or f'(x) is undefined. These are the only candidates for local maxima and minima. Think of it as: the function pauses (slope = 0) or does something abrupt (slope undefined).
Increasing / decreasing
A function is increasing where f'(x) > 0 and decreasing where f'(x) < 0. It changes from increasing to decreasing at a local maximum.
Absolute minimum (or maximum) on a closed interval
The smallest (or largest) value f actually takes on [a, b]. Found by evaluating f at every critical point in (a, b) and at both endpoints, then comparing.
Inflection point
A point where the concavity changes, i.e. where f''(x) changes sign. In simple terms, the curve switches from bending one way to bending the other.
Definite integral
The integral from a to b of f(x) dx, written with limits of integration. It gives a number (the net signed area under the curve between a and b).
Antiderivative
A function F(x) whose derivative is f(x). Finding antiderivatives is the core mechanic of integration.
U-substitution
A technique for undoing the chain rule in reverse. You set u equal to the inner function, compute du, and rewrite the integral in terms of u. Think of it as: find the composition hiding inside the integrand and simplify it.
f(x) = 1/(1 + x²). Find where f changes from increasing to decreasing
f'(x) = −2x/(1 + x²)²
f'(x) = 0 when x = 0. For x < 0, f' > 0 (increasing). For x > 0, f' < 0 (decreasing)
The function changes from increasing to decreasing at x = 0
Answer: 0
f(x) = x³ + x² − 8x + 5 on [−3, 2]
f'(x) = 3x² + 2x − 8 = (3x − 4)(x + 2), so critical points at x = 4/3 and x = −2
Evaluate f at x = −3, −2, 4/3, and 2
f(−3) = −27 + 9 + 24 + 5 = 11
f(−2) = −8 + 4 + 16 + 5 = 17
f(4/3) = 64/27 + 16/9 − 32/3 + 5 = −41/27
f(2) = 8 + 4 − 16 + 5 = 1
The absolute minimum value is −41/27
Answer: −41/27
f(x) = 3x⁵ − 10x³ + 5
f''(x) = 60x³ − 60x = 60x(x² − 1) = 60x(x − 1)(x + 1)
f'' changes sign at x = −1, 0, and 1, giving three inflection points
Answer: 3
Integral from 1 to 3 of 1/x² dx = integral of x⁻² dx
Antiderivative: −x⁻¹ = −1/x
Evaluate: [−1/x] from 1 to 3 = (−1/3) − (−1) = 2/3
Answer: 2/3
Integral from 0 to 3 of (e^(2x) + x³) dx
Antiderivative: (1/2)e^(2x) + x⁴/4
Evaluate: [(1/2)e⁶ + 81/4] − [(1/2)e⁰ + 0] = (1/2)e⁶ + 81/4 − 1/2
Simplify: (1/2)e⁶ + 81/4 − 2/4 = (1/2)e⁶ + 79/4
Answer: (1/2)e⁶ + 79/4
Integral from 0 to 1 of x²/(x³ + 1)² dx
Let u = x³ + 1, then du = 3x² dx, so x² dx = du/3
When x = 0, u = 1. When x = 1, u = 2
Integral becomes (1/3) · integral from 1 to 2 of u⁻² du
= (1/3)[−u⁻¹] from 1 to 2 = (1/3)(−1/2 + 1) = (1/3)(1/2) = 1/6
Answer: 1/6
Integral from e to e² of (ln x)²/x dx
Let u = ln x, then du = (1/x) dx
When x = e, u = 1. When x = e², u = 2
Integral becomes integral from 1 to 2 of u² du = [u³/3] from 1 to 2 = 8/3 − 1/3 = 7/3
Answer: 7/3
\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)Power rule for integration.
\int e^{ax} \, dx = \frac{1}{a} e^{ax} + CExponential integral.
\int \frac{1}{x} \, dx = \ln|x| + CLogarithmic integral.
U-substitution pattern: identify u = g(x), compute du = g'(x) dx, rewrite the integral in terms of u, integrate, substitute back (or change the limits if definite).
For absolute extrema on [a, b]: find all critical points in (a, b) where f'(x) = 0 or is undefined, evaluate f at each critical point and at both endpoints, then pick the largest and smallest values.
For inflection points: find where f''(x) = 0, then confirm f'' changes sign at each candidate.
Students often forget to check the endpoints when finding absolute extrema on a closed interval. Critical points alone are not enough; the endpoints can hold the absolute max or min.
A point where f''(x) = 0 is not automatically an inflection point. You must verify that f'' actually changes sign there. If it stays the same sign on both sides, there is no inflection.
When integrating x⁻², students sometimes use the power rule incorrectly, forgetting that the antiderivative of x⁻² is −x⁻¹, not x⁻¹. The sign matters.
In u-substitution, students forget to change the limits of integration when switching from x to u. If you do not change them, you must substitute back to x before evaluating.
⚠️ Absolute extrema on a closed interval is one of the most common exam problems. The method is formulaic: find critical points, evaluate f at each candidate, compare.
⚠️ Inflection point questions often ask "how many," not "where." Factor f'' completely and count sign changes.
⚠️ U-substitution appears in multiple forms. You may need to spot that x² dx is part of du when u = x³ + 1, or that (1/x) dx is du when u = ln x.
⚠️ The integral of e^(2x) is (1/2)e^(2x), not e^(2x). Forgetting the 1/a coefficient is one of the most common errors on integration problems.
True or false: To find the absolute minimum of f on [a, b], you only need to check where f'(x) = 0. (False. You must also check the endpoints a and b.)
Fill in the blank: An inflection point occurs where f''(x) changes ______. (sign)
True or false: The antiderivative of x⁻² is x⁻¹. (False. It is −x⁻¹.)
Fill in the blank: The integral of e^(2x) is ______ e^(2x) + C. (1/2)
True or false: In u-substitution on a definite integral, you can either change the limits to u-values or substitute back to x before evaluating. (True.)
Q: At what value of x does f(x) = 1/(1 + x²) change from increasing to decreasing?
A: f'(x) = −2x/(1 + x²)². This equals zero at x = 0. Since f' > 0 for x < 0 and f' < 0 for x > 0, the change happens at x = 0.
Q: Find the absolute minimum of f(x) = x³ + x² − 8x + 5 on [−3, 2].
A: f'(x) = 3x² + 2x − 8 = 0 gives x = 4/3 and x = −2. Evaluate: f(−3) = 11, f(−2) = 17, f(4/3) = −41/27, f(2) = 1. Minimum is −41/27.
Q: How many inflection points does f(x) = 3x⁵ − 10x³ + 5 have?
A: f''(x) = 60x³ − 60x = 60x(x − 1)(x + 1). This changes sign at x = −1, 0, and 1. Three inflection points.
Q: Evaluate the integral from 1 to 3 of 1/x² dx.
A: Antiderivative of x⁻² is −x⁻¹. Evaluate: −1/3 − (−1) = 2/3.
Q: Evaluate the integral from 0 to 1 of x²/(x³ + 1)² dx.
A: Let u = x³ + 1, du = 3x² dx. Limits become u = 1 to u = 2. Integral = (1/3) · [−u⁻¹] from 1 to 2 = (1/3)(1/2) = 1/6.
Q: Evaluate the integral from e to e² of (ln x)²/x dx.
A: Let u = ln x, du = (1/x) dx. Limits become u = 1 to u = 2. Integral = [u³/3] from 1 to 2 = 8/3 − 1/3 = 7/3.
The first derivative test connects back to differentiation rules: you need the product, quotient, or chain rule to find f' before you can analyse increasing/decreasing behaviour. U-substitution is the reverse of the chain rule, so fluency with the chain rule makes substitution much more intuitive. Inflection points become important again in curve sketching and in Calculus 2 when studying convergence of series.
First derivative test, second derivative test, critical points, absolute extrema, closed interval method, inflection points, concavity, definite integral, antiderivative, power rule integration, exponential integral, u-substitution, change of variable, MATH 104 Penn State, Calculus 1 optimisation, area under the curve