Single-Variable-Calculus · Unit 14 · Video 4 · Interactive Practice

Riemann Sums: the Definite Integral and a Year of Loans

IKey Formulas

Formula	Name	What you need
$\Delta x = \dfrac{b - a}{n}$	Width of each piece	Interval $[a, b]$ cut into $n$ equal pieces
$\displaystyle\sum_{i=1}^{n} f(c_i)\,\Delta x$	Riemann sum	A sample point $c_i$ in piece $i$
$\displaystyle\int_a^b f(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^{n} f(c_i)\,\Delta x$	Definite integral	Continuous $f$ ; any choice of the $c_i$
$\displaystyle\int_0^1 e^{r(1-t)}\,f(t)\,dt$	Total owed at the end of the year	Borrowing rate $f(t)$ , interest rate $r$ per year

Key Insight: The $dx$ in an integral is what the width $\Delta x$ becomes in the limit, so the integral carries the units of $f$ times the units of $x$ : dollars per year times years gives dollars.

IIVisualization 1 — Thin Rectangles, One Limit

For $x^2$ on $[0, 2]$ , left, right and midpoint staircases all close in on $\int_0^2 x^2\,dx = \tfrac{8}{3}$ .

Pieces n 4 Sample point

IIIVisualization 2 — A Year of Daily Loans

Total borrowed is a Riemann sum of rate times time step; with daily steps it is nearly $\int_0^1 f(t)\,dt$ .

Time step Piece i 45

💡 The rate here is $f(t) = 100{,}000 + 600{,}000\,t(1-t)^2$ dollars per year, whose integral over $[0, 1]$ is exactly $150{,}000$ dollars.

IVVisualization 3 — From Borrowed to Owed

A loan taken at time $t$ grows for $1 - t$ years, so the year-end debt weights $f(t)$ by $e^{r(1-t)}$ .

Interest rate r 0.05 Day i 45

VQuiz Questions

Problem 1 · A Right-Endpoint Riemann Sum

Given: $f(x) = x^2$ on $[1, 3]$ , cut into $n = 4$ equal pieces, with each sample point $c_i$ the right endpoint of its piece — find the Riemann sum $\sum_{i=1}^{4} f(c_i)\,\Delta x$ .

✅ Correct!

\Delta x = \tfrac{1}{2}

, the right endpoints are

1.5, 2, 2.5, 3

, and

\tfrac{1}{2}(2.25 + 4 + 6.25 + 9) = 10.75

❌ Those are the left endpoints.

1, 1.5, 2, 2.5

give

6.75

; the right endpoint of piece

i

c_i = 1 + \tfrac{i}{2}

, so the samples are

1.5, 2, 2.5, 3

❌ The heights are right, but each rectangle's area is height times width. Multiply

2.25 + 4 + 6.25 + 9 = 21.5

\Delta x = \tfrac{1}{2}

❌ That is the limit, not this sum.

\tfrac{26}{3}

is the area

\int_1^3 x^2\,dx

; with only

4

right-endpoint rectangles on a rising curve the sum overshoots it.

❌ Not quite. Find

\Delta x = \tfrac{3 - 1}{4}

, list the right endpoints, then add

f(c_i)\,\Delta x

Show solution

The width of each piece is

\Delta x = \frac{b - a}{n} = \frac{3 - 1}{4} = \frac{1}{2}.

The right endpoint of piece $i$ is $c_i = 1 + i\,\Delta x$ , so $c_1, c_2, c_3, c_4 = 1.5,\ 2,\ 2.5,\ 3$ . Then

\sum_{i=1}^{4} f(c_i)\,\Delta x = \tfrac{1}{2}\big(1.5^2 + 2^2 + 2.5^2 + 3^2\big) = \tfrac{1}{2}(2.25 + 4 + 6.25 + 9) = \tfrac{1}{2}(21.5) = 10.75.

Because $x^2$ rises across each piece, the right endpoint gives the tallest rectangle, so $10.75$ is larger than the exact area $\int_1^3 x^2\,dx = \tfrac{26}{3} \approx 8.67$ ; the left-endpoint sum, $6.75$ , is smaller.

Problem 2 · Keep Track of the Units

Given: A business borrows at a constant rate of $73,000 per year, with $t$ measured in years, and its bank adds up the daily pieces $\sum_{i=1}^{365} f\!\left(\tfrac{i}{365}\right)\Delta t$ — find the correct time step and the total borrowed over the year.

✅ Correct! Each day borrows $73,000 per year times

\tfrac{1}{365}

year, which is $200, and

365

such days make $73,000.

❌ That is a single term of the sum. $200 is one day's piece

f\!\left(\tfrac{i}{365}\right)\Delta t

; the total adds all

365

of them.

❌ The units do not match. The rate is in dollars per year, so

\Delta t

must be in years: one day is

\tfrac{1}{365}

year, not

1

. Using

1

inflates the total

365

-fold.

❌ Not quite. Check that (dollars per year) times (time step) comes out in dollars.

Show solution

Since $t$ is in years and $f$ is in dollars per year, the time step must be in years: one day is $\Delta t = \tfrac{1}{365}$ year. Then each term is

f\!\left(\tfrac{i}{365}\right)\Delta t = 73{,}000 \cdot \tfrac{1}{365} = 200 \text{ dollars},

and there are $365$ terms, so the total borrowed is $365 \cdot 200 = 73{,}000$ dollars. The units check: (dollars per year) $\times$ (years) $=$ dollars. The integral $\int_0^1 f(t)\,dt$ gives the same $73,000, the area of a rectangle of height $73{,}000$ and width $1$ .

Problem 3 · One Day's Loan at Year End

Given: On day $146$ (time $t = \tfrac{146}{365} = 0.4$ year) the business borrows $1,000. Interest is compounded continuously at $r = 0.05$ per year — find what that loan has grown to by the end of the year.

✅ Correct! The loan has

1 - 0.4 = 0.6

year left to grow, so it becomes

1000\,e^{0.05 \cdot 0.6} = 1000\,e^{0.03} \approx 1030.45

dollars.

❌ That loan was not out for a full year.

1000\,e^{0.05}

is a loan from

t = 0

; money borrowed at

t = 0.4

grows only for the remaining

1 - t

years.

❌ That uses the time already passed.

e^{0.05 \cdot 0.4}

counts the

0.4

year before the loan existed; the growth factor is

e^{r(1-t)}

, using the time still to come.

❌ That is simple interest.

1000(1 + 0.05 \cdot 0.6)

ignores compounding; continuous compounding multiplies by

e^{0.03}

, which is slightly more than

1.03

❌ Not quite. Use the growth factor

e^{r(1-t)}

with

t = 0.4

Show solution

Money borrowed at time $t$ has $1 - t$ years left in the year. Here $1 - 0.4 = 0.6$ year, which is $365 - 146 = 219$ days. With continuous compounding a principal $P$ grows to $P\,e^{r(1-t)}$ , so

1000\,e^{0.05\,(1 - 0.4)} = 1000\,e^{0.03} \approx 1000 \cdot 1.030455 \approx 1030.45 \text{ dollars}.

This growth factor $e^{r(1-t)}$ is exactly the weight that turns the total borrowed, $\int_0^1 f(t)\,dt$ , into the total owed, $\int_0^1 e^{r(1-t)} f(t)\,dt$ .

Problem 4 · Owed After Two Years

Given: The business borrows at rate $f(t)$ dollars per year during the first year only, $0 \le t \le 1$ , and repays nothing until the end of the second year, $t = 2$ , with interest compounded continuously at rate $r$ — which integral is the total owed at $t = 2$ ?

✅ Correct! A loan from time

t

grows for

2 - t

years before

t = 2

, so each piece

f(t)\,\Delta t

is weighted by

e^{r(2-t)}

❌ That is what is owed at $t = 1$ . The debt keeps growing through the second year, so each loan grows for

2 - t

years, not

1 - t

❌ That reverses the timing.

e^{rt}

would let the earliest loans grow least; a loan grows for the time remaining until repayment,

2 - t

❌ Interest compounds; it does not double the debt. Another year multiplies the year-one debt by

e^{r}

, not by

2

❌ Not quite. Ask how long the loan from time

t

grows before

t = 2

Show solution

Cut the first year into days of width $\Delta t = \tfrac{1}{365}$ . On day $i$ , at time $t = \tfrac{i}{365}$ , the business borrows $f\!\left(\tfrac{i}{365}\right)\Delta t$ dollars. That loan grows until $t = 2$ , which is $2 - \tfrac{i}{365}$ years, so at $t = 2$ it is worth

e^{r(2 - i/365)}\, f\!\left(\tfrac{i}{365}\right)\Delta t.

Adding over all $365$ days and letting the time step shrink gives

\sum_{i=1}^{365} e^{r(2 - i/365)}\, f\!\left(\tfrac{i}{365}\right)\Delta t \;\approx\; \int_0^1 e^{r(2-t)}\, f(t)\,dt.

Since $e^{r(2-t)} = e^{r}\,e^{r(1-t)}$ , this equals $e^{r}\int_0^1 e^{r(1-t)} f(t)\,dt$ : the amount owed at the end of the first year, grown for one more year.

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