Chapter 6.3, Problem 13E

To determine

The value of $y=-3x^2-1,[0,1]$

Answer to Problem 13E

The value of $x=\frac{1}{\sqrt{3}}$ .

Explanation of Solution

Given information: The function is $y=-3x^2-1,[0,1]$

Formula used:

  $av\left(f\right)=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$

Calculation:

If f is integrated on [a, b] its average (mean) value on [a, b] is

  $av\left(f\right)=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$

So,

  $\begin{array}{l}av\left(-3x^2-1\right)=\frac{1}{1-0}{\displaystyle \underset{0}{\overset{1}{\int }}\left(-3x^2-1\right)dx}\\ av\left(-3x^2-1\right)={\displaystyle \underset{0}{\overset{1}{\int }}\left(-3x^2-1\right)dx}\end{array}$

Now, calculate the integral by using NINT.

Calculator has a numerical integration capability, which will denote as NINT. In particular, using NINT $\left(f\left(x\right),x,a,b\right)$ to denote a calculator approximation of ${\displaystyle \underset{a}{\overset{b}{\int }}f}\left(x\right)dx$ . When write

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=\text{NINT}\left(f\left(x\right),x,a,b\right)$

The understanding that the right-hand side of the equation is an approximation of the left-hand side

Now, use above expression with this function

  $\begin{array}{l}f\left(x\right)=-3x^2-1\\ {\displaystyle \underset{0}{\overset{1}{\int }}\left(-3x^2-1\right)dx=\text{NINT}\left(-3x^2-1,x,0,1\right)}\\ =-2\end{array}$

So,

  $av\left(-3x^2-1\right)=-2$

Substitute the average value of $-3x^2-1$ in equation of y.

  $\begin{array}{l}-2=-3x^2-1\\ -3x^2=-1\\ x=\frac{1}{\sqrt{3}}\end{array}$

Conclusion:

The value of $x=\frac{1}{\sqrt{3}}$