4.3. The Area Problem http://www.ck12.org
The following example shows how we can use these to find the area.
Example 2:
Show that the upper and lower sums for the functionf(x) =x^2 ,fromx=0 tox= 1 ,approach the valueA= 1 / 3.
Solution:
LetPbe a partition ofnequal sub intervals over[ 0 , 1 ].We will show the result for the upper sums. By our definition
we have
T(P) =
n
∑ 1 Mi(xi−xi−^1 ) =M^1 (x^1 −x^0 )+M^2 (x^2 −x^1 )+...+Mn(xn−xn−^1 ).We note that each rectangle will have width^1 n,and lengths(^1 n)^2 ,(^2 n)^2 ,(^3 n)^2 ,...,(nn)^2 as indicated:
T(P) =
n
∑ 1 Mi(xi−xi−^1 ) =M^1 (x^1 −x^0 )+M^2 (x^2 −x^1 )+...+Mn(xn−xn−^1 )=^1 n( 1
n) 2
+^1 n( 2
n) 2
+^1 n( 3
n) 2
+...+^1 n(n
n) 2
=^1 n( 1
n) 2
( 12 + 22 + 32 +...+n^2 )=( 1
n^3)
( 12 + 22 + 32 +...+n^2 ) =( 1
n^3)(n(n+ 1 )( 2 n+ 1 )
6)
=
((n+ 1 )( 2 n+ 1 )
6 n^2)
.
We can re-write this result as:
(n+ 1 )( 2 n+ 1 )
6 n^2 =1
6
(n+ 1
n)( 2 n+ 1
n)
=^16
(
1 +^1 n)(
2 +^1 n)
.
We observe that as
n→+∞,^16(
1 +^1 n)(
2 +^1 n)
→^13.
We now are able to define the area under a curve as a limit.