To determine

To estimate:

The value of thegiven integral ∫02x3-3x+3dx

Answer

2≤ ∫02x3-3x+3dx≤10

Explanation

1) Concept:

Use comparison property 8 of the integral.

If m≤fx≤M for a≤x≤b, then mb-a≤∫abfxdx≤Mb-a

2) Given:

∫02x3-3x+3dx

3) Calculation:

Here,fx=x3-3x+3,  a=0,   b=2

Since this function is neither increasing nor decreasing over the given interval, we have to find its maximum and minimum over the given interval using thederivative test.

So, differentiate with respect to x

f'x=3x2-3=3x2-1=3(x+1)(x-1)

So f'x=0 at x=1 on the given interval [0, 2]

Therefore, check fx at x=0, 1, 2

To find the minimum and maximum value of fx

f0=03-30+3

f0=3

f1=13-31+3

f1=1

f2=23-32+3

f2=8-6+3

f2=5

So its minimum on 0, 2 is

m=f1=1

And its maximum on 0, 2 is

M=f2=5

Thus property 8 gives,

12-0≤  ∫02x3-3x+3dx≤52-0

2≤ ∫02x3-3x+3dx≤10

Conclusion:

The value of the integral is

2≤ ∫02x3-3x+3dx≤10