Question

Use Property 8 to estimate the value of the integral.

\(\int_0^2 {\left( {{x^3} - 3x + 3} \right)} dx\)

Short answer

The value of the integral \(\int_0^2 {\left( {{x^3} - 3x + 3} \right)} dx\) is \(2 \le \int_0^2 {\left( {{x^3} - 3x + 3} \right)} dx \le 10\)

Step-by-step solution

Property Used for the integral

Comparison property of the integral:

Let\(f\)be an integrable function with the limits\(a\)and\(b\)with\(a \le b\). If\(m \le f(x) \le M\)for\(a \le x \le b\), then,\(m(b - a) \le \int_a^b f (x)dx \le M(b - a)\).

Where, \(m\) is the minimum value of \(f\) and \(M\) is the maximum value of \(f\)."

find the critical numbers of \(f(x)\) in the open interval \((0,2)\)

To estimate the given integral, we need to find the absolute maximum/minimum values for\(f(x) = {x^3} - 3x + 3\)on the interval of integration, that is\((0,2)\)

Differentiate\(f(x)\)using the power rule

\(f'(x) = 3{x^{3 - 1}} - 3 \times 1 \times {x^{1 - 1}} + 0\)

\(f'(x) = 3{x^2} - 3\)

Since\(f\)and\({f^\prime }\)are polynomials, there are no critical numbers that arise due to the non-existence of the derivative.

\({f^'}(x) = 0 \Rightarrow3{x^2} - 3 = 0\)

Divide both sides by 3

\({x^2} - 1 = 0\)

Add 1 to both sides

\({x^2} = 1\)

Take the square root of both sides

\(x = \pm 1\)

Since, \(x = 1\) is the only critical number in the interval \((0,2)\)

Evaluate \(f(x)\) for the critical numbers as well as the endpoints

To find the absolute maximum/minimum values on a closed interval, evaluate the functional values at the critical numbers as well as the end points. Evaluate \(f(x)\) at \(x = 0,1,2\)

\(f(0) = {0^3} - 3 \times 0 + 3 = 3\)

\(f(1) = {1^3} - 3 \times 1 + 3 = 1\)

\(f(2) = {2^3} - 3 \times 2 + 3 = 5\)

Among the values above, the maximum is 5 and the minimum is 1

Therefore, using Property 8, we can write

\(1(2 - 0) \le \int_0^2 {\left( {{x^3} - 3x + 3} \right)} dx \le 5(2 - 0)\)

\(2 \le \int_0^2 {\left( {{x^3} - 3x + 3} \right)} dx \le 10\)

Hence, the value of the integral \(\int_0^2 {\left( {{x^3} - 3x + 3} \right)} dx\) is \(2 \le \int_0^2 {\left( {{x^3} - 3x + 3} \right)} dx \le 10\)