Basic Integration

Example 1. Prove f(x)≜xf(x)\triangleq xf(x)≜x
 is integrable on [a,b][a,b][a,b]
 and, in particular,
​∫abx dx=b2−a22.\displaystyle\int_a^b x\ \text{d}x=\dfrac{b^2-a^2}{2}.∫ab​x dx=2b2−a2​.
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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  over  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>0ε>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][a,b]
​ exceeds the lower Darboux integral by no more than ε\varepsilonε
​, i.e.
​inf⁡PU(f,P)⏟U(f)≤sup⁡PL(f,P)⏟L(f)+ε,\displaystyle \underbrace{\inf_{P} U(f,P)}_{U(f)} \leq \underbrace{\sup_{P}L(f,P)}_{L(f)} + \varepsilon,U(f)Pinf​U(f,P)​​≤L(f)Psup​L(f,P)​​+ε,
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​
where each PPP
 is a partition of [a,b].[a,b].[a,b].
 The second step is show U(f)=(b2−a2)/2U(f)=(b^2-a^2)/2U(f)=(b2−a2)/2
.
Step 1. For each n∈Nn\in\mathbb{N}n∈N
, define the regular partition Pn={x0,x1,…,xn}P_n=\{ x_0, x_1,\ldots,x_n \}Pn​={x0​,x1​,…,xn​}
, where
​xi≜a+i⋅b−an,for all i∈[n].x_i \triangleq a + i\cdot \dfrac{b-a}{n}, \quad \text{for all $i\in[n]$}.xi​≜a+i⋅nb−a​,for all i∈[n].
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​
so that xi+1−xi=(b−a)/nx_{i+1}-x_i = (b-a)/nxi+1​−xi​=(b−a)/n
 for each i.i.i.
Then
​xi+1=f(xi+1)=sup⁡x∈[xi,xi+1]f(x).\displaystyle x_{i+1}=f(x_{i+1})=\sup_{x\in[x_i,x_{i+1}]} f(x).xi+1​=f(xi+1​)=x∈[xi​,xi+1​]sup​f(x).
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Hence
​U(f,Pn)=∑i=0n−1(sup⁡x∈[xi,xi+1]f(x))⋅(xi+1−xi)=b−an⋅∑i=0n−1xi+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}.U(f,Pn​)=i=0∑n−1​(x∈[xi​,xi+1​]sup​f(x))⋅(xi+1​−xi​)=nb−a​⋅i=0∑n−1​xi+1​.
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Since
​∑i=0n−1(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},i=0∑n−1​(i+1)=i=1∑n​i=2n(n+1)​,
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we deduce
​U(f,Pn)=b−an⋅∑i=0n−1(a+(i+1)⋅b−an)=b−an(na+n+12⋅(b−a))=b2−a22+(b−a)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}U(f,Pn​)​=nb−a​⋅i=0∑n−1​(a+(i+1)⋅nb−a​)=nb−a​(na+2n+1​⋅(b−a))=2b2−a2​+2n(b−a)2​.​
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By analogous argument,
​L(f,Pn)=b2−a22−(b−a)22n.\displaystyle L(f,P_n)=\dfrac{b^2-a^2}{2}-\dfrac{(b-a)^2}{2n}.L(f,Pn​)=2b2−a2​−2n(b−a)2​.
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Combining our results reveals
​U(f,Pn)−L(f,Pn)=(b−a)2n,for all n∈N.\displaystyle U(f,P_n)-L(f,P_n)=\dfrac{(b-a)^2}{n}, \quad \text{for all}\ n\in\mathbb{N}.U(f,Pn​)−L(f,Pn​)=n(b−a)2​,for all n∈N.
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By the Archimedean property of R\mathbb{R}R
, there is N∈NN\in\mathbb{N}N∈N
 such that
​(b−a)2≤Nε  ⟹  (b−a)2N≤ε.(b-a)^2 \leq N\varepsilon \quad\implies\quad\dfrac{(b-a)^2}{N}\leq\varepsilon.(b−a)2≤Nε⟹N(b−a)2​≤ε.
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Hence
​U(f)−L(f)≤U(f,PN)−L(f,PN)=(b−a)2N≤ε.\displaystyle U(f)-L(f)\leq U(f,P_N)-L(f,P_N)=\dfrac{(b-a)^2}{N}\leq \varepsilon.U(f)−L(f)≤U(f,PN​)−L(f,PN​)=N(b−a)2​≤ε.
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Since ε\varepsilonε
 was chosen arbitrarily, we deduce U(f)≤L(f)U(f)\leq L(f)U(f)≤L(f)
, which the first result follows.
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Step 2. By construction,
​b2−a22−(b−a)22n=L(f,Pn)≤L(f)≤U(f)≤U(f,Pn)=b2−a22+(b−a)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}2b2−a2​−2n(b−a)2​​=L(f,Pn​)≤L(f)≤U(f)≤U(f,Pn​)=2b2−a2​+2n(b−a)2​.​
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As the Archimedean property also implies 1/n→01/n\rightarrow 01/n→0
, applying the squeeze lemma yields
​∫abx dx=U(f)=b2−a22.\displaystyle\int_a^b x\ \text{d}x=U(f)=\dfrac{b^2-a^2}{2}.∫ab​x dx=U(f)=2b2−a2​.
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​■\blacksquare■
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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 7x4−4x2+87x^4-4x^2+87x4−4x2+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.
Darboux Sums
Optimization
Last modified 3mo ago