Integral of 3x from 0 to 4 from the definition, with upper and lower sums, step by step

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The problem

Show from the definition that this integral exists, and compute it with lower and upper sums. The picture is drawn with 4 pieces.

\int_{0}^{4} 3 x \, dx

The result

\int_{0}^{4} 3 x \, dx = 24

The function is integrable: over $n$ equal pieces the upper and lower sums differ by exactly $\frac{48}{n}$ , and both converge to $24$ .

checked 18 lines of this solution were evaluated with exact arithmetic before it was shown to you. A solution with a false line is never served.

Picture 1 of 2Step 1: Say exactly what "integrable" and "the integral" mean

The graph over the interval cut into 4 equal pieces. On each piece the lower rectangle reaches up to the least value and the upper rectangle to the greatest value; the strip between them is that piece's share of the gap. With

n = 4

the lower sum is

18

and the upper sum is

30

; the integral,

24

, lies between them.

Say exactly what "integrable" and "the integral" mean

Why this step

"Area under a curve" is a picture, not a definition: a curved region has no formula for its area until we make one. The definition builds that meaning out of the one area everybody agrees on, the rectangle, and everything we do afterwards is forced by it.

Definition

Partition, lower sum, upper sum

A partition $P$ of $[a, b]$ is a finite list of cut points $a = x_0 < x_1 < \cdots < x_n = b$ . It chops the interval into pieces $[x_{i-1}, x_i]$ of width $\Delta x_i = x_i - x_{i-1}$ . On each piece let $m_i$ be the least value of $f$ and $M_i$ the greatest value of $f$ . The lower sum is $L(P, f) = \sum_{i=1}^{n} m_i \, \Delta x_i$ : rectangles that stay UNDER the graph. The upper sum is $U(P, f) = \sum_{i=1}^{n} M_i \, \Delta x_i$ : rectangles that COVER the graph. The symbol $\sum$ (capital sigma) means "add up": $\sum_{i=1}^{n}$ says let $i$ run through $1, 2, \ldots, n$ and add every term.

\text{Claim: } f(x) = 3 x \text{ is integrable on } [0, 4] \text{ and } \int_{0}^{4} 3 x \, dx = 24

Two things to prove: the vice closes, and the number it traps is

24

n = 4

the lower sum is

18

and the upper sum is

30

; the integral,

24

, lies between them.

Choose the partitions: n equal pieces

Why this step

The definition asks only that SOME partition works for each $\varepsilon$ . We are free to pick the friendliest family there is. Equal pieces make every width the same number, which lets that width come out of the sums as a common factor. Unequal pieces would prove the same thing with far uglier algebra.

\Delta x = \frac{4}{n}

Cut

[0, 4]

, which has length

4 - 0 = 4

, into

n

equal pieces. Each piece has this width. checked

x_i = \frac{4 i}{n} \qquad (i = 0, 1, 2, \ldots, n)

Start at the left end

0

and take

i

steps of width

\Delta x

. Check the two ends:

i = 0

gives

x_0 = 0

and

i = n

gives

x_n = 0 + 4 = 4

Call this partition $P_n$ . There is one for every natural number $n$ ; larger $n$ means thinner pieces.

n = 4

the lower sum is

18

and the upper sum is

30

; the integral,

24

Integral from the definition with upper and lower sums

Say exactly what "integrable" and "the integral" mean

Choose the partitions: n equal pieces

Find the least and greatest value on each piece

Measure the gap between the upper and lower sums

Compute the upper sum exactly

Let n grow, and trap the integral

Watch the vice close