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How do you find the average value of the function for #f(x)=e^x, -1<=x<=1#?

Answer 1

Isaiah Allen

# bar (f)(x) = 1/2 \ (e-1/e) ~~ 1.1752 #

By definition, the average value of a function #f(x)# over a domain #[a,b]# is given by:

# bar (f)(x) = 1/(b-a) \ int_a^b \ f(x) \ dx #

So, for the given function, #f(x)=e^x# with #-1 le x le 1#, we have:

# bar (f)(x) = 1/(1-(-1)) \ int_(-1)^(1) \ e^x \ dx #

# \ \ \ \ \ \ \ = 1/2 \ [e^x]_(-1)^(1) #

# \ \ \ \ \ \ \ = 1/2 \ (e-e^(-1)) #

# \ \ \ \ \ \ \ = 1/2 \ (e-1/e) #

# \ \ \ \ \ \ \ ~~ 1.1752 #

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Answer 2

Adam Adkins

To find the average value of the function ( f(x) = e^x ) over the interval ([-1, 1]), you can use the formula:

[ \text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) , dx ]

where ( a ) and ( b ) are the limits of integration, which in this case are ( -1 ) and ( 1 ) respectively.

So,

[ \text{Average value} = \frac{1}{1 - (-1)} \int_{-1}^{1} e^x , dx ]

[ = \frac{1}{2} \left[ e^x \right]_{-1}^{1} ]

[ = \frac{1}{2} \left( e^1 - e^{-1} \right) ]

[ = \frac{1}{2} \left( e - \frac{1}{e} \right) ]

[ = \frac{e}{2} - \frac{1}{2e} ]

[ \approx 1.543 ]

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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.

How do you find the average value of the function for #f(x)=e^x, -1&lt;=x&lt;=1#?

How do you find the average value of the function for #f(x)=e^x, -1<=x<=1#?