Question

The area A between the graph of the function:

g(t) = 4 - (4/t^2)

and the t-axis over the interval [1, x] is:

A(x) = ∫[1, x] (4 - (4/t^2)) dt

a) Find the horizontal asymptote of the graph g.

I believe the horizontal asymptote of graph g is g(t) = 4.

b) Integrate to find A as a function of x. Does the graph of A have a horizontal asymptote? Explain.

I went ahead and computed the integral which I got
4t + (4/t) + C
But am I writing this correctly? I'm not exactly sure how question b wants me to write the answer out. Also how would I find if the graph of A have a horizontal asymptote?

Any help is greatly appreciated!

Answer 1

You have the right idea, but you are working a definite integral, so there is no constant C. Instead, evaluate A(t) at the interval endpoints:

A(x) = (4t + 4/t)[1,x]
= (4x + 4/x)-(4*1 + 4/1)
= 4x + 4/x - 8