Average mean and antiderivative?

Question

f(x)=10(x^1/2)+8 or (10 sqrt(x) +8) on the interval [3,10]
what is the average/mean slope on this interval.

[4sqrt(t^3)-3t]/sqrt(t^3) x(9)=7
what is the antiderivative

Thanks so much

the x(9)=7 is not part of the equation it is something that is given in the probelm. Thanks

Answer 1

I take the "mean slope" to mean
change in function value
divided by
change in x value.

So work out f(10) - f(3) and then divide by 7.

√(t^3) = t^(3/2)

Answer 2

okay, let's look at the antiderivative first. So put

dx = 4 [sqrt(t^3) - 3t] / sqrt (t^3) dt.
Then
x = 4 integral [sqrt(t^3) - 3t] / sqrt(t^3) dt
x = 4 integral [ 1 - t^(1 - 3/2)] dt
x = 4 integral [ 1 - t^(-1/2) ] dt
x = 4(t - 2 t^(1/2)) + c
x = 4t - 8 sqrt(t) + c.
Using the initial condition x(9) = 7, put t = 9 to get
7 = 36 - 24 + c
therefore c = -5.
This yields x = 4t - 8 sqrt(t) - 5.