http://www.ck12.org Chapter 4. Integration
4.4 Definite Integrals
Learning Objectives
- Use Riemann Sums to approximate areas under curves
- Evaluate definite integrals as limits of Riemann Sums
Introduction
In the Lesson The Area Problem we defined the area under a curve in terms of a limit of sums.
A=nlim→+∞S(P) =nlim→+∞T(P)where
S(P) =
n
∑ 1 mi(xi−xi−^1 ) =m^1 (x^1 −x^0 )+m^2 (x^2 −x^1 )+...+mn(xn−xn−^1 ),T(P) =n
∑ 1 Mi(xi−xi− 1 ) =M 1 (x 1 −x 0 )+M 2 (x 2 −x 1 )+...+Mn(xn−xn− 1 ),S(P),andT(P)were examples ofRiemann Sums. In general, Riemann Sums are of form∑ni= 1 f(x∗i) 4 xwhere each
x∗iis the value we use to find the length of the rectangle in theithsub-interval. For example, we used the maximum
function value in each sub-interval to find the upper sums and the minimum function in each sub-interval to find the
lower sums. But since the function is continuous, we could have used any points within the sub-intervals to find the
limit. Hence we can define the most general situation as follows:
Definition
Iffis continuous on[a,b],we divide the interval[a,b]intonsub-intervals of equal width with 4 x=b−na. We
letx 0 =a,x 1 ,x 2 ,...,xn=bbe the endpoints of these sub-intervals and letx∗ 1 ,x∗ 2 ,...,x∗nbeanysample points
in these sub-intervals. Then thedefinite integralofffromx=atox=bis∫b
a f(x)dx=nlim→∞n
∑i= 1 f(x∗i)^4 x.Example 1:
Evaluate the Riemann Sum forf(x) =x^3 fromx=0 tox=3 usingn=6 sub-intervals and taking the sample points
to be the midpoints of the sub-intervals.