What is the average value of the function #h(r) = 3 / (1+r)^2 # on the interval #[1,6]#?

Answer 1

Charles Anctil

It is #3/14#

The mean value is provided by

#h_(avg)=1/(b-a)int_a^b h(r)dr=> h_(avg)=1/(6-1)int_1^6 3/(1+r)^2dr=> h_(avg)=1/515/14=3/14#

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Answer 2

Paisley Johnson

To find the average value of a function ( h(r) ) over an interval ([a,b]), you can use the formula:

[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} h(r) , dr ]

For the given function ( h(r) = \frac{3}{(1+r)^2} ) on the interval ([1,6]), the average value is:

[ \text{Average value} = \frac{1}{6-1} \int_{1}^{6} \frac{3}{(1+r)^2} , dr ]

Now, integrate the function over the interval:

[ \int_{1}^{6} \frac{3}{(1+r)^2} , dr = \left[-\frac{3}{1+r}\right]_{1}^{6} = \left(-\frac{3}{7}\right) - \left(-\frac{3}{2}\right) = \frac{15}{14} ]

So, the average value of ( h(r) ) on the interval ([1,6]) is ( \frac{15}{14} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.