Question

Use Property 8 to estimate the value of the integral.

\(\int_0^2 {\sqrt {{x^3} + 1} } dx\)

Short answer

The value of the integral \(\int_0^2 {\sqrt {{x^3} + 1} } dx\) is \(2 \le \int_0^2 {\sqrt {{x^3} + 1} } dx \le 6\)

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 \(f\)."

Calculate the first derivative of the function \(f\).

First, calculate the first derivative of the function\(f\).

\({f^\cent}(x) = \frac{{3{x^2}}}{{2\sqrt {{x^3} + 1} }}\)

Since functions\(x \mapsto 3{x^2}\)and\(x \mapsto 2\sqrt {{x^3} + 1} \)are both nonnegative on the interval\(\left( {0,2} \right),\)we conclude that\(f'\)is nonnegative on\((0,2)\)and positive on\((0,2)\).

Hence, the function\(f\)is an increasing on\((0,2)\)and the minimum\(m\)is on the left boundary 0, and the maximum\(M\)is on the right boundary 2.

Then,

\(m: = f(0) = 1 \le f(x) \le 3 = f(2) = :M,\quad \forall x \in (0,2)\)

Using the comparison property of the integral

Now, by the comparison property of the integral we have

\(m(b - a) \le \int_a^b f (x)dx \le M(b - a)\)

where\(a = 0\)and\(b = 2\). Thus

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

\(2 \le \int_0^2 {\sqrt {{x^3} + 1} } dx \le 6\)

Hence, the value of \(\int_0^2 {\sqrt {{x^3} + 1} } dx\) is \(2 \le \int_0^2 {\sqrt {{x^3} + 1} } dx \le 6\)