AP Course

AP Calculus AB/BC

Study the complete College Board sequence for AP Calculus AB and BC, from limits through infinite series.

Choose an official unit and topic to open its lecture, concept check, or focused practice.

Lessons
1.1 Introducing Calculus: Can Change Occur at an Instant?1.2 Defining Limits and Using Limit Notation1.3 Estimating Limit Values from Graphs1.4 Estimating Limit Values from Tables1.5 Determining Limits Using Algebraic Properties of Limits1.6 Determining Limits Using Algebraic Manipulation1.7 Selecting Procedures for Determining Limits1.8 Determining Limits Using the Squeeze Theorem1.9 Connecting Multiple Representations of Limits1.10 Exploring Types of Discontinuities1.11 Defining Continuity at a Point1.12 Confirming Continuity over an Interval1.13 Removing Discontinuities1.14 Connecting Infinite Limits and Vertical Asymptotes1.15 Connecting Limits at Infinity and Horizontal Asymptotes1.16 Working with the Intermediate Value Theorem (IVT)2.1 Defining Average and Instantaneous Rates of Change at a Point2.2 Defining the Derivative of a Function and Using Derivative Notation2.3 Estimating Derivatives of a Function at a Point2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist2.5 Applying the Power Rule2.6 Derivative Rules: Constant, Sum, Difference, and Constant Multiple2.7 Derivatives of cos x, sin x, e^x, and ln x2.8 The Product Rule2.9 The Quotient Rule2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions3.1 The Chain Rule3.2 Implicit Differentiation3.3 Differentiating Inverse Functions3.4 Differentiating Inverse Trigonometric Functions3.5 Selecting Procedures for Calculating Derivatives3.6 Calculating Higher-Order Derivatives4.1 Interpreting the Meaning of the Derivative in Context4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration4.3 Rates of Change in Applied Contexts Other Than Motion4.4 Introduction to Related Rates4.5 Solving Related Rates Problems4.6 Approximating Values of a Function Using Local Linearity and Linearization4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms5.1 Using the Mean Value Theorem5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points5.3 Determining Intervals on Which a Function Is Increasing or Decreasing5.4 Using the First Derivative Test to Determine Relative (Local) Extrema5.5 Using the Candidates Test to Determine Absolute (Global) Extrema5.6 Determining Concavity of Functions over Their Domains5.7 Using the Second Derivative Test to Determine Extrema5.8 Sketching Graphs of Functions and Their Derivatives5.9 Connecting a Function, Its First Derivative, and Its Second Derivative5.10 Introduction to Optimization Problems5.11 Solving Optimization Problems5.12 Exploring Behaviors of Implicit Relations6.1 Exploring Accumulations of Change6.2 Approximating Areas with Riemann Sums6.3 Riemann Sums, Summation Notation, and Definite Integral Notation6.4 The Fundamental Theorem of Calculus and Accumulation Functions6.5 Interpreting the Behavior of Accumulation Functions Involving Area6.6 Applying Properties of Definite Integrals6.7 The Fundamental Theorem of Calculus and Definite Integrals6.8 Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation6.9 Integrating Using Substitution6.10 Integrating Functions Using Long Division and Completing the Square6.11 Integrating Using Integration by Parts6.12 Using Linear Partial Fractions6.13 Evaluating Improper Integrals6.14 Selecting Techniques for Antidifferentiation7.1 Modeling Situations with Differential Equations7.2 Verifying Solutions for Differential Equations7.3 Sketching Slope Fields7.4 Reasoning Using Slope Fields7.5 Approximating Solutions Using Euler’s Method7.6 Finding General Solutions Using Separation of Variables7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables7.8 Exponential Models with Differential Equations7.9 Logistic Models with Differential Equations8.1 Finding the Average Value of a Function on an Interval8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts8.4 Finding the Area Between Curves Expressed as Functions of x8.5 Finding the Area Between Curves Expressed as Functions of y8.6 Finding the Area Between Curves That Intersect at More Than Two Points8.7 Volumes with Cross Sections: Squares and Rectangles8.8 Volumes with Cross Sections: Triangles and Semicircles8.9 Volume with Disc Method: Revolving Around the x- or y-Axis8.10 Volume with Disc Method: Revolving Around Other Axes8.11 Volume with Washer Method: Revolving Around the x- or y-Axis8.12 Volume with Washer Method: Revolving Around Other Axes8.13 The Arc Length of a Smooth, Planar Curve and Distance Traveled9.1 Defining and Differentiating Parametric Equations9.2 Second Derivatives of Parametric Equations9.3 Finding Arc Lengths of Curves Given by Parametric Equations9.4 Defining and Differentiating Vector-Valued Functions9.5 Integrating Vector-Valued Functions9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions9.7 Defining Polar Coordinates and Differentiating in Polar Form9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve9.9 Finding the Area of the Region Bounded by Two Polar Curves10.1 Defining Convergent and Divergent Infinite Series10.2 Working with Geometric Series10.3 The nth Term Test for Divergence10.4 Integral Test for Convergence10.5 Harmonic Series and p-Series10.6 Comparison Tests for Convergence10.7 Alternating Series Test for Convergence10.8 Ratio Test for Convergence10.9 Determining Absolute or Conditional Convergence10.10 Alternating Series Error Bound10.11 Finding Taylor Polynomial Approximations of Functions10.12 Lagrange Error Bound10.13 Radius and Interval of Convergence of Power Series10.14 Finding Taylor or Maclaurin Series for a Function10.15 Representing Functions as Power Series
Quizzes
Practice Problems AP formula notes, graph references, and practice sets will be added here.

AP Calculus AB/BC · Unit 6 · Topic 6.9

Integrating Using Substitution

Reverse the chain rule by replacing an inner expression and its differential.

1. Topic Focus

Interpret definite integrals as accumulated change, connect sums to integrals, apply both Fundamental Theorems, and select antiderivative techniques.

This topic: Reverse the chain rule by replacing an inner expression and its differential.

2. Key Relationship

\(u=g(x),\quad du=g'(x)dx\)

Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.

3. Visual Connection

x-integralf(g(x))times g'(x) dxbounds: a, bchooseu = g(x)replacedu = g'(x)dxchange boundsu-integralf(u) dubounds:g(a), g(b)one variable at every stageintegrate and verify
Substitution changes the entire integration languageReplace the inner expression and differential together, and transform both limits whenever a definite integral remains in the new variable.

4. Worked Example

For ∫2x cos(x²)dx, let u=x² to obtain sin(x²)+C.

Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.

5. Concept Development

Substitution reverses the chain rule

The derivative of \(F(g(x))\) is \(F'(g(x))g'(x)\). Integration by substitution recognizes that chain-rule structure and replaces the inner expression with a simpler variable.

\(u=g(x),\quad du=g'(x)dx,\quad \int f(g(x))g'(x)dx=\int f(u)du.\)

Choose the inner expression that simplifies the integral

Useful candidates commonly appear inside a power, radical, exponential, logarithm, denominator, or trigonometric function. Differentiate the candidate and compare it with the remaining factors.

The differential replaces a complete factor

After setting \(u=g(x)\), write \(du=g'(x)dx\). Treat \(g'(x)dx\) as one package and determine exactly what constant multiple is present in the original integrand.

Constant-factor adjustments are allowed

If \(du=6x,dx\) but the integral contains \(x,dx\), then \(x,dx=\frac16du\). Constants may move outside the integral; variable factors may not.

Every part of the integrand must use one variable

A completed substitution contains only \(u\) and \(du\). If an \(x\) remains, rewrite it using the substitution or choose a more effective substitution.

Indefinite integrals return to the original variable

Integrate in \(u\), substitute \(u=g(x)\) back into the answer, and include \(+C\). The final antiderivative should normally be expressed in the variable from the original problem.

Definite integrals require corresponding new bounds

\(\int_a^b f(g(x))g'(x)dx=\int_{g(a)}^{g(b)}f(u)du.\)

Evaluate the substitution at each original endpoint before integrating in \(u\).

Do not mix variables and bounds

A \(u\)-integral must have \(u\)-bounds. If the antiderivative is converted back to \(x\), use the original \(x\)-bounds instead. Either route is valid, but combining them is not.

A decreasing substitution may reverse the new bounds

If \(g(a)>g(b)\), the transformed lower bound is larger than the upper bound. Keep that order and allow orientation to handle the sign rather than silently rearranging the bounds.

Power patterns are common substitution targets

\(\int[g(x)]^n g'(x)dx=\frac{[g(x)]^{n+1}}{n+1}+C,\qquad n\ne-1.\)

The logarithmic pattern uses a function over itself

\(\int\frac{g'(x)}{g(x)}dx=\ln|g(x)|+C.\)

The absolute value is required unless the domain guarantees that \(g(x)>0\).

Exponential patterns preserve the inner expression

\(\int e^{g(x)}g'(x)dx=e^{g(x)}+C.\)

For \(a^{g(x)}\), divide by \(\ln a\) after substitution.

Trigonometric substitutions reverse chain-rule derivatives

For example, \(\int\cos(g(x))g'(x)dx=\sin(g(x))+C\), while a sine pattern introduces a negative cosine. Signs should be checked by differentiation.

Algebra may be needed after choosing u

Sometimes the substitution lets you solve for a remaining factor. If \(u=x+1\), then \(x=u-1\); if \(u=x^2+1\), then \(x^2=u-1\). Rewrite before integrating.

Substitution is not always the right method

The presence of a composite function alone is insufficient. A matching derivative factor must be available or obtainable by a constant adjustment or valid algebraic rewrite.

Differentiate to verify an indefinite result

Applying the chain rule to the final expression should reproduce the original integrand exactly. This detects missing constants, incorrect signs, and incomplete back-substitution.

A reliable substitution workflow

Select \(u\), compute \(du\), adjust constants, rewrite everything in \(u\), transform bounds when present, integrate, return to \(x\) only when needed, and verify.

Common errors

Frequent errors include choosing an inner function without its derivative, dropping a coefficient, leaving mixed variables, using original bounds on a \(u\)-integral, adding \(+C\) to a definite value, and changing reversed bounds without changing sign.

6. Detailed Worked Example and Error Check

Example 1: Reverse a polynomial chain rule.

\(\int6x(3x^2+5)^4dx=\int u^4du=\frac{(3x^2+5)^5}{5}+C,\)

where \(u=3x^2+5\) and \(du=6x\,dx\).

Example 2: Adjust a missing constant. Let \(u=x^3+4\), so \(du=3x^2dx\). Then

\(\int x^2\sqrt{x^3+4}\,dx=\frac13\int u^{1/2}du=\frac29(x^3+4)^{3/2}+C.\)

Example 3: Recognize the logarithmic pattern.

\(\int\frac{4x-1}{2x^2-x+7}dx=\ln(2x^2-x+7)+C.\)

The numerator is exactly the derivative of the denominator, which is always positive.

Example 4: Track a trigonometric sign. With \(u=\cos x\) and \(du=-\sin xdx\),

\(\int\sin x\cos^4x\,dx=-\int u^4du=-\frac15\cos^5x+C.\)

Example 5: Change the bounds of a definite integral. For \(u=x^2+1\), the bounds \(x=0,2\) become \(u=1,5\):

\(\int_0^2x(x^2+1)^3dx=\frac12\int_1^5u^3du=\left[\frac{u^4}{8}\right]_1^5=78.\)

Example 6: Preserve decreasing-bound orientation. Let \(u=\cos x\), so \(du=-\sin xdx\). The bounds \(0\) and \(\pi/2\) become \(1\) and \(0\):

\(\int_0^{\pi/2}\sin x\,e^{\cos x}dx=-\int_1^0e^u du=\int_0^1e^u du=e-1.\)

Example 7: Rewrite a remaining variable. Set \(u=x+1\), so \(x=u-1\) and \(dx=du\). Then

\(\begin{aligned}\int\frac{x}{\sqrt{x+1}}dx&=\int\frac{u-1}{\sqrt u}du\\&=\frac23u^{3/2}-2u^{1/2}+C\\&=\frac23(x+1)^{3/2}-2\sqrt{x+1}+C.\end{aligned}\)

7. AP Reasoning Routine

Identify the accumulating quantity and units, preserve bounds, choose a valid integration technique, and check answers by differentiation.

  • Identify the representation and requested quantity.
  • State the rule or theorem and verify its conditions.
  • Keep exact values until the final requested approximation.
  • Interpret sign, units, interval, and context.
AP Checkpoint

Use substitution and show the choice of \(u\), the differential, and any transformed bounds.
(a) Evaluate \(\int8x(4x^2-3)^6dx\).
(b) Evaluate \(\int\frac{x^2}{x^3+2}dx\).
(c) Evaluate \(\int e^{5x-1}dx\).
(d) Evaluate \(\int\cos(3x)dx\).
(e) Evaluate \(\int\frac{\cos x}{2+\sin x}dx\).
(f) Evaluate \(\int_1^3\frac{2x}{x^2+4}dx\) using transformed bounds.
(g) Evaluate \(\int_0^1 3x^2\sqrt{x^3+1}\,dx\).
(h) Evaluate \(\int_0^{\pi/2}\cos x(1+\sin x)^2dx\).
(i) A student writes \(u=x^2+1\) and then \(\frac12\int_0^2u^3du\). Explain the notation error for the original integral \(\int_0^2x(x^2+1)^3dx\).
(j) Evaluate \(\int x^3\sqrt{x^2+1}\,dx\) by rewriting the remaining \(x^2\) after substitution.

Check the solution

(a) Let \(u=4x^2-3\), so \(du=8x\,dx\). The result is \((4x^2-3)^7/7+C\).
(b) Let \(u=x^3+2\), so \(du=3x^2dx\). The result is \(\frac13\ln|x^3+2|+C\).
(c) Let \(u=5x-1\), so \(dx=du/5\). The result is \(\frac15e^{5x-1}+C\).
(d) With \(u=3x\), the result is \(\frac13\sin(3x)+C\).
(e) Let \(u=2+\sin x\), so \(du=\cos xdx\). The result is \(\ln|2+\sin x|+C\).
(f) Let \(u=x^2+4\); the bounds become 5 and 13. Thus \(\int_5^{13}du/u=\ln(13/5)\).
(g) Let \(u=x^3+1\); the bounds become 1 and 2. The value is \(\int_1^2u^{1/2}du=\frac23(2\sqrt2-1)\).
(h) Let \(u=1+\sin x\); the bounds become 1 and 2. The value is \(\int_1^2u^2du=7/3\).
(i) After changing to \(u\), the limits must also change from \(x=0,2\) to \(u=1,5\). The correct transformed integral is \(\frac12\int_1^5u^3du\).
(j) Let \(u=x^2+1\), so \(x^2=u-1\) and \(x\,dx=du/2\). Then \(\frac12\int(u-1)u^{1/2}du=\frac15u^{5/2}-\frac13u^{3/2}+C\), giving \(\frac15(x^2+1)^{5/2}-\frac13(x^2+1)^{3/2}+C\).