Definite Integrals 235Example 1
Use a midpoint Riemann sum with three subdivisions of equal length to find the approxi-
mate value of
∫ 6
0 x(^2) dx.
Δx=
6 − 0
3
=2, f(x)=x^2midpoints arex=1, 3, and 5.
∫ 6
0x^2 dx≈ f(1)Δx+f(3)Δx+f(5)Δx=1(2)+9(2)+25(2)≈ 70Example 2
Using the limit of the Riemann sum, find
∫ 5
13 xdx.
Usingnsubintervals of equal lengths, the length of an interval
Δxi=5 − 1
n=
4
n;xi= 1 +(
4
n)
i
∫ 513 xdx= lim
maxΔxi→ 0∑ni= 1f(ci)Δxi.Letci=xi; maxΔxi→ 0 ⇒n→∞.
∫ 513 xdx=nlim→∞∑ni= 1f(
1 +
4 i
n)(
4
n)
=nlim→∞∑ni= 13
(
1 +
4 i
n)(
4
n)=nlim→∞12
n∑ni= 1(
1 +
4 i
n)
=nlim→∞12
n(
n+4
n[
n(
n+ 1
2)])=nlim→∞12
n(n+ 2 (n+ 1 ))=lim
n→∞12
n( 3 n+ 2 )=lim
n→∞(
36 +24
n)
= 36Thus,
∫ 513 xdx=36.Properties of Definite Integrals
- Iffis defined on [a,b], and the limit lim
maxΔxi→ 0
∑n
i= 1f(xi)Δxiexists, thenfis integrable
on [a,b].- Iff is continuous on [a,b], then fis integrable on [a,b].
Iff(x),g(x), andh(x) are integrable on [a,b], then