Basic Integration
Example 1. Prove
f(x)xf(x)\triangleq x
is integrable on
[a,b][a,b]
and, in particular,
abx dx=b2a22.\displaystyle\int_a^b x\ \text{d}x=\dfrac{b^2-a^2}{2}.
Ex 1: Comment
Hint
Visual
Formal Write-Up
Follow-Up
Here we must work from the definitions of integration rather that simply apply the fundamental theorem of calculus and power rule.
Choose a sequence of regular partitions and compute the upper/lower Darboux sums for an arbitrary index of this sequence. It will match, up to a diminishing term, the sought value.
Integration estimate of
f(x)f(x)
over
[a,b][a,b]
with errors highlighted. Upper Darboux sums are shown with regular partitions.
To verify integrability, it suffices to show the upper and lower Darboux integrals are equal (Step 1). Then we show these integrals equal the desired value (Step 2). Let
ε>0\varepsilon>0
be given. Since lower integrals do not exceed upper integrals, the first step may be completed by showing the upper Darboux integral on
[a,b][a,b]
​ exceeds the lower Darboux integral by no more than
ε\varepsilon
​, i.e.
infPU(f,P)U(f)supPL(f,P)L(f)+ε,\displaystyle \underbrace{\inf_{P} U(f,P)}_{U(f)} \leq \underbrace{\sup_{P}L(f,P)}_{L(f)} + \varepsilon,
where each
PP
is a partition of
[a,b].[a,b].
The second step is show
U(f)=(b2a2)/2U(f)=(b^2-a^2)/2
.
Step 1. For each
nNn\in\mathbb{N}
, define the regular partition
Pn={x0,x1,,xn}P_n=\{ x_0, x_1,\ldots,x_n \}
, where
xia+iban,for all i[n].x_i \triangleq a + i\cdot \dfrac{b-a}{n}, \quad \text{for all $i\in[n]$}.
so that
xi+1xi=(ba)/nx_{i+1}-x_i = (b-a)/n
for each
i.i.
Then
xi+1=f(xi+1)=supx[xi,xi+1]f(x).\displaystyle x_{i+1}=f(x_{i+1})=\sup_{x\in[x_i,x_{i+1}]} f(x).
Hence
U(f,Pn)=i=0n1(supx[xi,xi+1]f(x))(xi+1xi)=bani=0n1xi+1.\displaystyle U(f,P_n)=\sum_{i=0}^{n-1} \left(\sup_{x\in[x_i,x_{i+1}]} f(x)\right)\cdot (x_{i+1}-x_i)=\dfrac{b-a}{n}\cdot\sum_{i=0}^{n-1}x_{i+1}.
Since
i=0n1(i+1)=i=1ni=n(n+1)2,\displaystyle\sum_{i=0}^{n-1}( i+1)=\sum_{i=1}^n i=\dfrac{n(n+1)}{2},
we deduce
U(f,Pn)=bani=0n1(a+(i+1)ban)=ban(na+n+12(ba))=b2a22+(ba)22n.\displaystyle \begin{array}{rl}U(f,P_n) &=\displaystyle\dfrac{b-a}{n}\cdot\sum_{i=0}^{n-1}\left(a+(i+1)\cdot\dfrac{b-a}{n}\right)\\&=\dfrac{b-a}{n}\left(na+\dfrac{n+1}{2}\cdot(b-a)\right)\\ & = \dfrac{b^2-a^2}{2}+\dfrac{(b-a)^2}{2n}.\end{array}
By analogous argument,
L(f,Pn)=b2a22(ba)22n.\displaystyle L(f,P_n)=\dfrac{b^2-a^2}{2}-\dfrac{(b-a)^2}{2n}.
Combining our results reveals
U(f,Pn)L(f,Pn)=(ba)2n,for all nN.\displaystyle U(f,P_n)-L(f,P_n)=\dfrac{(b-a)^2}{n}, \quad \text{for all}\ n\in\mathbb{N}.
By the Archimedean property of
R\mathbb{R}
, there is
NNN\in\mathbb{N}
such that
(ba)2Nε    (ba)2Nε.(b-a)^2 \leq N\varepsilon \quad\implies\quad\dfrac{(b-a)^2}{N}\leq\varepsilon.
Hence
U(f)L(f)U(f,PN)L(f,PN)=(ba)2Nε.\displaystyle U(f)-L(f)\leq U(f,P_N)-L(f,P_N)=\dfrac{(b-a)^2}{N}\leq \varepsilon.
Since
ε\varepsilon
was chosen arbitrarily, we deduce
U(f)L(f)U(f)\leq L(f)
, which the first result follows.
Step 2. By construction,
b2a22(ba)22n=L(f,Pn)L(f)U(f)U(f,Pn)=b2a22+(ba)22n.\begin{array}{rl}\dfrac{b^2-a^2}{2}-\dfrac{(b-a)^2}{2n}&=L(f,P_n)\\&\leq L(f)\\ &\leq U(f) \\ &\leq U(f,P_n) \\ &=\dfrac{b^2-a^2}{2}+\dfrac{(b-a)^2}{2n}.\end{array}
As the Archimedean property also implies
1/n01/n\rightarrow 0
, applying the squeeze lemma yields
abx dx=U(f)=b2a22.\displaystyle\int_a^b x\ \text{d}x=U(f)=\dfrac{b^2-a^2}{2}.
\blacksquare
Typically, it's easiest to work with regular partitions. These often admit nice formulas for the "error" term in the integrals. The sum formulas used here only work in special cases and, as the proof was already rather long, one may expect to not have to do too complicated of integral proofs (e.g. with
7x44x2+87x^4-4x^2+8
) directly from definitions.
In the formal write-up, we deduced the amount of error from our integral over estimate analytically. However, we could arrive at the same result geometrically, as shown in the "Visual" tab.
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