4.4. Definite Integrals http://www.ck12.org
Solution:
If we partition the interval[ 0 , 3 ]inton=6 equal sub-intervals, then each sub-interval will have length^3 − 60 =^12 .So
we have 4 x=^12 and
R 6 =
6
∑ 1 f(x∗i)^4 x=f(^0.^25 )^4 x+f(^0.^75 )^4 x+f(^1.^25 )^4 x+f(^1.^75 )^4 x+f(^2.^25 )^4 x+f(^2.^75 )^4 x=( 1
64
)( 1
2
)
+
( 27
64
)( 1
2
)
+
( 125
64
)( 1
2
)
+
( 343
64
)( 1
2
)
+
( 729
64
)( 1
2
)
+
( 1331
64
)( 1
2
)
=^255664 = 39. 93.
Now let’s compute the definite integral using our definition and also some of our summation formulas.
Example 2:
Use the definition of the definite integral to evaluate∫ 03 x^3 dx.
Solution:
Applying our definition, we need to find
∫ 3
0 x(^3) dx=lim
n→∞
n
i∑= 1 f(x∗i)^4 x.
We will use right endpoints to compute the integral. We first need to divide[ 0 , 3 ]intonsub-intervals of length
4 x=^3 −n^0 =^3 n.Since we are using right endpoints,x 0 = 0 ,x 1 =^3 n,x 2 =^6 n,...xi=^3 ni.
So∫ 03 x^3 dx=limn→∞∑ni= 1 f(^3 ni)(^3 n) =limn→∞^3 n∑ni= 1 (^3 ni)^3 =limn→∞^3 n∑in= 1 (^27 n 3 )i^3 =limn→∞^81 n 4 ∑ni= 1 i^3.
Recall that∑n 1 i^3 =
[n(n+ 1 )
2
] 2
.By substitution, we have∫ 3
0 x(^3) dx=lim
n→∞=
81
n^4[n(n+ 1 )
2] 2
=nlim→∞^814[
1 +^1 n] 2
→^814 asn→∞.Hence