Use the Fundamental Theorem of Calculus to find the average value of f(x)=e^0.4x between x=0 and x=2.

Find the average value on the graph. Finally give the exact x value for which f(x)=the average value.
Answer for the x value: ?

I keep getting 40ln(20e^0.8-20) and it is wrong.

from my sketch it would be

∫ e^.4x dx from 0 to 2
= [ (1/.4) e^.4x dx from 0 to 2
= (1/.4) e^(.4)(2) - (1/.4) e^0
= 2.5( e^.8 - 1)
= appr 3.064

How does ln possibly enter the picture in your answer ?

the average value is 1/2 the area above

To find the average value of a function using the Fundamental Theorem of Calculus, you need to follow these steps:

Step 1: Find the definite integral of the function over the given interval.
In this case, the integral of f(x) = e^0.4x from x = 0 to x = 2 is:

∫[0,2] e^0.4x dx

Step 2: Evaluate the definite integral.
To evaluate the definite integral, you can use the power rule for integration. The power rule states that the integral of x^n with respect to x is (1/n+1) * x^(n+1) + C. Applying this rule, we can evaluate the integral as follows:

∫[0,2] e^0.4x dx = (1/0.4) * e^0.4x ∣[0,2]
= 2.5 * ( e^0.8 - e^0 )
= 2.5 * ( e^0.8 - 1 )

Therefore, the definite integral of f(x) over the interval [0,2] is 2.5 * ( e^0.8 - 1 ).

Step 3: Calculate the length of the interval.
The length of the interval [0,2] is given by the difference between the upper and lower limit: 2 - 0 = 2.

Step 4: Divide the integral by the length of the interval.
To find the average value, divide the definite integral by the length of the interval:

Average value = ( 2.5 * ( e^0.8 - 1 ) ) / 2
= 1.25 * ( e^0.8 - 1 )

The average value of f(x) = e^0.4x between x = 0 and x = 2 is 1.25 * ( e^0.8 - 1 ).

Finally, to find the exact x value for which f(x) equals the average value, set f(x) equal to the average value and solve for x.

1.25 * ( e^0.8 - 1 ) = e^0.4x

To solve this equation, we can divide both sides by 1.25:

e^0.8 - 1 = ( e^0.4x ) / 1.25

Next, we can take the natural logarithm (ln) of both sides:

ln(e^0.8 - 1) = ln(( e^0.4x ) / 1.25)

x = ln((e^0.8 - 1)/1.25)

Therefore, the exact x value for which f(x) equals the average value is x = ln((e^0.8 - 1)/1.25).