To determine
Part (a):
To find:
The lower estimate for ∫1030f(x)dx
Answer
The lower estimate: L5=-64
Explanation
1) Concept:
To find the lower estimate by using the formula of the Riemann sum for left end points.
2) Formula:
The Riemann sum for the left endpoints:Ln=∑i=1nf(xi-1)∆x
where ∆x=b - an,n is the number of the subintervals.
3) Given:
4) Calculation:
Since f(x) is increasing,L5 ≤ ∫1030fxdx ≤R5.
Here, n=5,a=10 and b=30
So, the width of the subintervals is
∆x=b-an
∆x=30-105
∆x=205=4
Use ∆x=4 to form the subintervals.
10, 14, 14, 18, 18, 22, 22, 26, [26, 30]
Therefore, the left endpoints are x0=10, x1=14, x2=18, x3=22, x4=26
Using the formula of the lower estimate: L5=∑i=15f(xi-1)∆x
L5=∆x(fx0+fx1+fx2+fx3+fx4)
L5=∆x(f10+f14+f18+f22+f26)
From the given table,
f10=-12, f14=-6, f18=-2, f22=1, f26=3
Substitute these values and ∆x=4.
L5=4(-12-6-2+1+3)
L5=4-16=-64
Therefore,
The lower estimate: L5=-64
Conclusion:
The lower estimate: L5=-64
To determine
Part (b):
To find:
The upper estimates for ∫1030f(x)dx
Answer
The upper estimate: R5=16
Explanation
1) Concept:
To find the upper estimate by using the Riemann sum formula for the right endpoints.
2) Formula:
The Riemann sum for the right endpoints: Rn=∑i=1nf(xi)∆x
where ∆x=b - an,n is the number of the subintervals.
3) Given:
4) Calculation:
Since f(x) is increasing,L5 ≤ ∫1030fxdx ≤R5.
Here, n=5,a=10, and b=30.
So, the width of the subintervals is
∆x=b-an
∆x=30-105
∆x=205=4
Use ∆x=4 to form the subintervals.
10, 14, 14, 18, 18, 22, 22, 26, [26, 30]
Therefore, the right endpoints are x1=14, x2=18, x3=22, x4=26 and x5=30
By using the formula of the upper estimate: R5=∑i=15f(xi)∆x
R5=∆x(fx1+fx2+fx3+fx4+fx5)
R5=∆x(f14+f18+f22+f26+f 30)
From the given table,
f14=-6, f18=-2, f22=1, f26=3, f30=8
Substitute these all values and ∆x=4.
R5=4(-6-2+1+3+8)
R5=44=16
Therefore,
The upper estimate: R5=16
Conclusion:
The upper estimate: R5=16