How do we find the exact area under a curve? Approximate with rectangles, then take n→∞.
Δx=nb−a
Left, Right, and Midpoint Sums
Type
Sample Point
Formula
Left (Ln)
Left endpoint xi
Worked Example
Approximate∫04x2dxwithn=4using Left Riemann Sum.
Δx=1. Left endpoints: 0,1,2,3.
L4=f(0)+f(1)+f(2)
The exact answer is 364≈21.33, so L4= is an .
Over- and Underestimates — Essential AP Skill
The relationship between the function’s behavior and the estimate type determines over/under.
Function Behavior
Left Sum
Right Sum
Midpoint
Trapezoidal
Increasing
Under
Over
—
Over
Decreasing
Over
Under
—
Under
Riemann Sum Computations 🎯
Trapezoidal Rule
The Trapezoidal Rule averages the Left and Right sums:
Tn
Trapezoidal Rule from a Table 🎯
Given the table:
x
0
2
5
8
10
f(x)
3
7
11
6
4
Classify each approximation. 🔍
Let f be a positive, increasing, concave-up function on [a,b].
Compute a Riemann Sum. ✍️
Key Takeaways — Part 1
Concept
Key Formula
Subinterval width
Δx=(b−a)/n
Left Sum
Use left endpoints
Right Sum
Use right endpoints
Midpoint Sum
Use midpoint of each subinterval
Trapezoidal
Part 2: Definite Integral Definition
∫ Definite Integrals
Part 2 of 7 — The Definite Integral
From Riemann Sums to Exact Area
The definite integral is the limit of a Riemann sum as n→∞:
Part 3: Properties of Integrals
∫ Definite Integrals
Part 3 of 7 — Properties of Integrals
Essential Properties
∫ab[c
Part 4: Fundamental Theorem of Calculus — Part 1
∫ Definite Integrals
Part 4 of 7 — FTC Part 1
The Fundamental Theorem — Part 1
dxd∫
Part 5: Fundamental Theorem of Calculus — Part 2
∫ Definite Integrals
Part 5 of 7 — FTC Part 2 & Net Change
The Evaluation Theorem (FTC Part 2)
∫ab
Part 6: Mixed Integration Problems
∫ Definite Integrals
Part 6 of 7 — Problem-Solving Workshop
Combining All Tools
This part brings together everything: Riemann sums, FTC, properties, and applications.
Strategy Guide
Problem Type
Key Approach
Evaluate ∫abf(x)dx
Find antiderivative, apply FTC Part 2
Part 7: Comprehensive Review
∫ Definite Integrals — Comprehensive Review
Part 7 of 7 — Final Assessment
Complete Summary
Concept
Key Formula
Riemann Sum
∑f(xi∗)Δx
Trapezoidal Rule
∑i=0n−1f(xi)Δx
Right (Rn)
Right endpoint xi+1
∑i=1nf(xi)Δx
Midpoint (Mn)
Midpoint xˉi
∑i=1nf(2xi−1+
+
f(3)=
0+
1+
4+
9=
14
14
underestimate
Concave Up
—
—
Under
Over
Concave Down
—
—
Over
Under
Key Concept: For increasing functions, left rectangles miss the top-right corner (under), while right rectangles include extra area (over). The reverse for decreasing.
AP Tip: The AP Exam loves asking "Is this an over- or underestimate?" You MUST justify by citing whether f is increasing/decreasing (for L/R) or concave up/down (for M/T).
=
2Δx
[f(x0)+2f(x1)+2f(x2)+⋯+2f(xn−1)+f(xn)]
Pattern: First and last values appear ONCE; all middle values are DOUBLED.
Equal vs. Unequal Subintervals
With unequal subintervals (common on AP tables), apply the trapezoid formula to each pair:
T=2Δx1[f(x0)+f(x1)]+2Δx2[f(x1)+f(x2)]+⋯
Worked Example
Trapezoidal approximation of∫04x2dxwithn=4:
T4=2L4+R4=214+30=22
Exact: 64/3≈21.33. Since x2 is concave up, the Trapezoidal sum overestimates. ✓
AP Tip: The trapezoidal rule with table data is one of the most common AP FRQ questions. Practice with unequal subintervals!
2Δx[f(x0)+2f(x1)+⋯+f(xn)]
Over/Under (L/R)
Depends on increasing/decreasing
Over/Under (M/T)
Depends on concavity
Up Next: Part 2 — The Definite Integral.
∫abf(x)dx=n→∞limi=1∑nf(xi∗)Δx
Signed Area
Region
Sign
Between curve and x-axis, curve ABOVE axis
Positive
Between curve and x-axis, curve BELOW axis
Negative
Signed area=∫abf(x)dxTotal area=∫ab∣f(x)∣dx
Key Concept: The definite integral gives signed area, not total area. The AP Exam tests this distinction frequently!
Geometric Evaluation
Some integrals can be evaluated using geometry instead of antiderivatives:
Shape
Integral
Value
Rectangle
∫035dx
5×3=15
Triangle
∫042xdx
2
Trapezoid
∫03(2x+1)dx
Semicircle
∫−rrr2−x
Verify:∫03(2x+1)dx=[x ✓
Worked Example — Semicircle
∫−339−x2 = area of semicircle with =
AP Tip: If you see r2−x2, think semicircle! No antiderivative needed.
Definite Integral Concepts 🎯
Even and Odd Function Shortcuts
Odd: ∫−aaf(x)dx=0Even: ∫−aaf(x)dx=2∫0af(x)dx
Type
Definition
Examples
Integral on [−a,a]
Odd
f(−x)=−f(x)
Worked Example
∫−11(x4+x3)
Split: even
=2∫01x4dx+0=
AP Tip: Check for symmetry BEFORE computing! It can save significant time on the exam.
Evaluate Definite Integrals 🎯
Interpret each integral. 🔍
Geometric Evaluation ✍️
Key Takeaways — Part 2
Concept
Key Point
Definition
∫abfdx=limn→∞∑f(xi∗)Δx
Signed area
Above axis = +, below = −
Total area
$\int
Odd functions
Integral = 0 on symmetric intervals
Even functions
2× integral from 0 to a
Geometry
Use triangles, trapezoids, semicircles
Up Next: Part 3 — Properties of Integrals.
f
(
x
)]
d
x
=
c
∫ab
f
(
x
)
d
x
∫ab[f(x)±g(x)]dx=∫abf(x)dx±∫abg(x)dx
Property
Formula
Interpretation
Constant Multiple
∫abcfdx=c∫abfdx
Factor constants out
Sum/Difference
∫ab(f±g)dx=∫f±
Additivity
∫abfdx+∫b
Reversal
∫abfdx=−∫b
Zero Width
∫aafdx=0
No interval = no area
Comparison
f≥g⇒∫f≥∫g
Bigger function = bigger integral
Key Concept: Integrals are linear — they respect addition and scalar multiplication. This is used constantly on the AP Exam!
Worked Examples — Given-Value Problems
A common AP question gives you known integral values and asks you to find others.
Given:∫05f(x)dx=10, ∫05g(x)dx=3, ∫03f(x)dx=7.
Find
Work
Answer
∫05[2f(x)−3g(x)]d
AP Tip: Don't forget that ∫abkdx=k(b−a) for a constant . Many students miss the constant term!
Apply Integral Properties 🎯
Given: ∫05f(x)dx=10 and ∫05g(x)dx=3.
Average Value of a Function
favg=b−a1∫abf(x)dx
Interpretation: The average height of the function over [a,b].
Geometric meaning:favg⋅(b−a)=∫. The rectangle with height has the SAME area as the region under the curve.
Worked Example
Find the average value of f(x)=x2 on [0,3].
favg=3−0
AP Tip: The Mean Value Theorem for Integrals guarantees there exists a c∈[a,b] where f(c)=favg (if is continuous). They may ask you to find this .
Average Value 🎯
Apply Properties 🔍
Given: ∫06f(x)dx=15, ∫06g(x)dx=7, ∫04f(x)dx=9.
Compute with Properties ✍️
Key Takeaways — Part 3
Property
Formula
Linearity
Constants factor out, sums split
Additivity
∫ab+∫bc=∫ac
Reversal
Swap limits → flip sign
Average value
b−a1∫abf
Constant integral
∫abkdx=k(b−a)
Up Next: Part 4 — FTC Part 1.
ax
f
(
t
)
d
t
=
f
(
x
)
In words: Differentiation undoes integration. If you integrate f and then differentiate, you get f back.
With the Chain Rule
If the upper limit is a function g(x):
dxd∫ag(x)f(t)dt=f(g(x))⋅g′(x)
All Variations at a Glance
Situation
Formula
Key Step
Upper limit = x
dxd∫axf(t)dt=f(x)
Direct application
Upper limit = g(x)
f(g(x))⋅g′(x)
Chain Rule
Lower limit = x
dxd∫
Both limits are functions
f(g(x))g′(x)−f(h(x))h
AP Tip: FTC Part 1 with the Chain Rule is tested almost every year on the AP Exam. Master this!
Worked Examples
Example 1:dxd∫2xt3dt=x3 ✓ (direct)
Example 2:dxd∫0x
g(x)=x2, g′(x)=2x:
Example 3:dxd∫x5e
Reverse: =−dxd∫5x
Example 4 (Both limits):dxd∫2xx
Split: ∫0x3costdt−∫
=cos(x3)⋅3x2−
FTC Part 1 🎯
Accumulation Functions
F(x)=∫axf(t)dt is an accumulation function: it measures how much f has "accumulated" from a to x.
Connecting f and F
About f
About F=∫axfdt
F(a)=∫aaf(t)
Key Concept: If they give you the graph of f′, you can determine the behavior of f using this same table (since f=∫f). This is one of the most common AP graph-analysis questions.
Accumulation Functions 🎯
Let F(x)=∫0xf(t)dt where f is continuous.
FTC Part 1 — Match the derivative. 🔍
Compute a specific value. ✍️
Key Takeaways — Part 4
Concept
Formula
FTC Part 1 (basic)
dxd∫axf(t)dt=f(x)
FTC Part 1 (chain)
dxd∫
Variable in lower limit
Reverse limits → negative sign
F(a)=0
Accumulation starts at 0
f>0⇒F increasing
f<0⇒F decreasing
Up Next: Part 5 — FTC Part 2 & Net Change.
f
(
x
)
d
x
=
F
(
b
)
−
F
(
a
)
where
F′
=
f
Notation:F(x)ab or [F(x)]ab=F(b)−F(a)
Quick Evaluation Examples
Integral
Antiderivative
Evaluation
∫023x2dx
x3
8−0=8
∫1ex1dx
∫01exdx
e
∫0π/2cosxdx
sinx
Key Fact: You can use ANY antiderivative — so always choose C=0 for simplicity.
Evaluate Using FTC Part 2 🎯
Net Change Theorem
∫abf′(x)dx=f(b)−f(a)
The integral of a rate of change gives the NET CHANGE in the original quantity.
Applications Table
Quantity
Rate
∫ab(rate)dt gives...
Position s(t)
Velocity
Displacement vs Total Distance
Displacement=∫abv
Concept
Formula
Includes direction?
Displacement
∫vdt
Yes (can be negative)
Total Distance
$\int
v
AP Tip: "How far" = total distance (∫∣v∣). "What is the displacement" or "change in position" = ∫v. Read the question carefully!
Worked Example — Displacement vs Distance
A particle has v(t)=t2−4 on [0,3].
Displacement:∫03(t2−4)dt=[3t
The particle is 3 units to the LEFT of where it started.
Total Distance:v(t)=0 at t=2. Split at the zero:
∫02∣t
=[4t−3t
Key Concept: To compute ∫∣v∣dt, find where v=0, split the integral, and negate v on intervals where v.
Net Change Theorem 🎯
Interpret each integral. 🔍
Net Change Problem ✍️
Key Takeaways — Part 5
Concept
Formula
FTC Part 2
∫abf=F(b)−F(a)
Net Change
∫abf′=f(b)−
Displacement
∫abvdt (signed)
Total Distance
$\int_a^b
Position update
s(b)=s(a)+∫abvdt
Up Next: Part 6 — Problem-Solving Workshop.
dxd∫ag(x)f(t)dt
FTC Part 1 (+ Chain Rule if needed)
Given integral values
Use linearity and additivity properties
Table data
Trapezoidal rule (unequal subintervals)
Rate → total change
Net Change Theorem: ∫abf′=f(b)−f(a)
Even/odd symmetry
Simplify before computing
Absolute value
Split at zeros, negate on negative intervals
AP Tip: On FRQs, always show your setup (the integral expression) before evaluating. Setup points are awarded separately from answer points.
AP-Style Mixed Problems — Set 1 🎯
Absolute Value Integrals — Step by Step
To evaluate ∫ab∣f(x)∣dx:
Find where f(x)=0 (the zeros)
Determine sign of f on each subinterval
Split the integral at each zero
Negate f on intervals where f<0
Worked Example
∫04∣x−2∣dx
x−2=0 at x=2.
On [0,2]: x−2<0, so ∣x−2∣
=∫02(2−
=(4−2)+(8−8)−(2−4)=
Geometric shortcut:∣x−2∣ forms a V-shape — two right triangles each with base 2 and height 2. Area = 2×21. ✓
Mixed Problems — Set 2 🎯
Classify each problem type. 🔍
Trapezoidal Rule from a Table ✍️
t (min)
0
3
7
10
R(t) (gal/min)
4
6
10
8
Key Takeaways — Part 6
Problem Type
Go-To Tool
Evaluate definite integral
FTC Part 2
Differentiate an integral
FTC Part 1
Given values problems
Properties (linearity, additivity)
Rate → amount
Net Change Theorem
Table data
Trapezoidal Rule
Absolute value
Split at zeros
Up Next: Part 7 — Comprehensive Review.
2Δx[f(x0)+2f(x1)+⋯+f(xn)]
Definite Integral
limn→∞∑f(xi∗)Δx
FTC Part 1
dxd∫axf(t)dt=f(x)
FTC Part 1 (chain)
dxd∫ag(x)f(t)dt=f(g(x))⋅g′(x)
FTC Part 2
∫abf(x)dx=F(b)−F(a)
Net Change
∫abf′(x)dx=f(b)−f(a)
Average Value
favg=b−a1∫abf(x)dx
Common AP Exam Mistakes
Mistake
Consequence
Forgetting Chain Rule on FTC Part 1
Missing the g′(x) factor
Confusing displacement with distance
Using ∫v when asked for $\int
Not splitting at zeros for $\int
f
Forgetting ∫kdx=k(b−a) for constants
Missing the constant term
Wrong Trapezoidal with unequal widths
Using equal Δx when widths vary
Not reversing limits when x is in lower bound
Sign error
Final Assessment — Set 1 🎯
Final Assessment — Set 2 🎯
AP-Style Comprehensive 🔍
Final Challenge ✍️
Definite Integrals — Complete! ✅
You have mastered:
Skill
Parts
Riemann Sums (L, R, M, T)
Part 1
Over/Underestimate analysis
Part 1
Signed area & geometry
Part 2
Even/odd symmetry
Part 2
Properties & average value
Part 3
FTC Part 1 (+ Chain Rule)
Part 4
Accumulation functions
Part 4
FTC Part 2 & Net Change
Part 5
Displacement vs distance
Part 5
Mixed problem solving
Parts 6-7
AP Exam Checklist
✅ Can evaluate definite integrals with FTC Part 2
✅ Can differentiate integrals with FTC Part 1 (+ Chain Rule)
✅ Can use properties to compute from given values
✅ Can apply Trapezoidal Rule to table data
✅ Can distinguish displacement from total distance
✅ Can analyze accumulation functions from graphs of f
FTC connects differentiation and integration — they are INVERSE operations!