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Grade 12 / High 3 (age 17-18)

Definite Integrals & Series

Definite Integrals & Series

Slice Thin, Sum Up, Get Area
6
👀 See It
①Split the interval into n parts and add all rectangle areas
②As n grows the staircase hugs the curve
③In the limit, the sum is exactly the definite integral (area) — this is the Riemann sum
The Riemann Sum
The heart of the Riemann sum
limn→∞ 1n ∑k=1n f(kn) = ∫01 f(x) dx
1/n is the width (Δx), k/n is the position (x) — the limit of the sum is the integral
General interval [a,b]
limn→∞ ∑k=1n f(a + (b−a)kn)·b−an = ∫ab f(x) dx
Rectangles of width (b−a)/n at position a+(b−a)k/n
Turning the Limit Sum Into an Integral
🔁 Substitution rule
①Replace k/n with x
②Replace 1/n with dx
③The sum over k=1..n becomes the integral from 0 to 1
④Even messy limit sums collapse to one integral by this rule
Compute It Directly
Example 1
Find limn→∞ 1n ∑k=1n kn.
1
Replace k/n with x and 1/n with dx.
= ∫01 x dx
2
Evaluate the integral.
= [x22]01 = 12
▸ 1/2
See the k/n inside as the variable x and the leading 1/n as dx — done.
Example 2
Find limn→∞ 1n ∑k=1n (kn)2.
1
Convert to an integral by the same rule.
= ∫01 x2 dx
2
Evaluating gives
= [x33]01 = 13
▸ 1/3
(k/n)² is x², 1/n is dx — powers do not change the rule.
Wrap-up
Key result
limn→∞ 1n ∑k=1n f(kn) = ∫01 f(x) dx
The sum of infinitely many rectangles = the definite integral (area)
2020 CSAT Math (Calculus) type, adapted
Find limn→∞ 1n ∑k=1n √(k/n).
①1/2
②2/3
③3/4
④1
⑤Diverges
▸ ② 2/3
1
Replacing k/n→x and 1/n→dx gives an integral.
= ∫01 √x dx
2
Integrate x1/2.
= [23 x3/2]01 = 23
🎯 Exam Points
①Seeing 1/n·Σf(k/n) means ∫₀¹f(x)dx at once
②The k/n→x, 1/n→dx substitution is key
③Roots and powers do not change the rule
④General interval: width (b−a)/n, position a+(b−a)k/n
⑤If the 1/n in front is missing, it is not a Riemann sum
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