To find the area under a curve, start by cutting the interval into thin strips, filling them with rectangles and adding them all up.
The more rectangles you use, the closer the jagged staircase hugs the curve, and the limit as their number grows without bound is the definite integral.
So by turning k/n inside the sum into x and the 1/n in front into dx, a complicated limit becomes a single integral.
In the picture, raising the number of rectangles n brings their sum toward the value of the integral, 1/3.
Area from adding thin slices
seegongsik.com
f(x) = x² on [0, 1] filled with n rectangles (each as tall as the function at the right end of its strip); the sum Σ and the limit 1/3 are shown at the top
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👀 What to read in the picture
①Split [0, 1] into n equal parts and add the areas of the rectangles
②Using right-end heights, the sum is always a little more than 1/3
③As n grows, the staircase hugs the curve and the sum approaches 1/3
④Finding area this way is called a Riemann sum
The Riemann Sum
The key formula
limn→∞1n ∑k=1n f(kn) = ∫01 f(x) dx
1/n is the width (Δx) and k/n the position (x); the limit of the sum is the definite integral
A general interval [a, b]
limn→∞ ∑k=1n f(a + (b−a)kn)·b−an = ∫ab f(x) dx
The sum of rectangles of width (b − a)/n at positions a + (b − a)k/n
Turning a limit of sums into an integral
🔁 The substitution rule
①Replace k/n by x
②Replace 1/n by dx
③The sum from k = 1 to n becomes the integral from 0 to 1
④For an interval other than [0, 1], use width (b − a)/n and position a + (b − a)k/n
Worked examples
Example 1
Find limn→∞1n ∑k=1nkn.
1
Replace k/n by x and 1/n by dx.
= ∫01 x dx
2
Evaluate the definite integral.
= [x22]01 = 12
▸ 1/2
It adds up rectangles of width 1/n and height k/n, so it equals the area of the triangle under y = x from 0 to 1, which is 1/2.
Example 2
Find limn→∞1n ∑k=1n (kn)2.
1
Turn it into a definite integral by the same rule.
= ∫01 x2 dx
2
Evaluate the definite integral.
= [x33]01 = 13
▸ 1/3
(k/n)² becomes x² and 1/n becomes dx. The picture shows exactly this sum, which approaches 1/3 as n grows.
Summary
Key point
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 by x and 1/n by dx turns it into a definite integral.
= ∫01 √x dx
2
Integrate √x = x1/2.
= [23 x3/2]01 = 23
🎯 Exam Points
①(1/n) Σ f(k/n) turns into the integral of f(x) from 0 to 1
②The key is k/n → x and 1/n → dx
③The rule is the same with roots or powers
④For a general interval, use width (b − a)/n and position a + (b − a)k/n
⑤1/n becomes dx only when the interval has length 1