Question

find the value(s) of c guaranteed by the Mean Value Theorem for Integrals for the function over the given interval.

Answer 1

Remember that

If f is continuous over [a,b] and differentiable over (a,b), then there exists c∈(a,b) such that

`f^(\prime)(c)=(f(b)-f(a))/(b-a)`

In this problem, we have the function

`f(x)=(9)/(x^3)`

over the interval [1,3]

so

f(a)=f(1)=9/(1)^3=9

f(b)=f(3)=9/(3)^3=1/3

substitute

`f^(\prime)(c)=((1)/(3)-9)/(3-1)=(-(26)/(3))/(2)=-(26)/(6)=-(13)/(3)`

Find out the first derivative f'(x)

`f^(\prime)(x)=-(27)/(x^4)`

Equate the first derivative to -13/3

`\begin{gathered} -(27)/(x^4)=-(13)/(3) \\ \\ x^4=(27*3)/(13) \\ \\ x=1.58 \end{gathered}`

therefore

The value of c is 1.58 (rounded to two decimal places)