Calculus 2 question

Question

Determine all values of T for which the curves defined by y=Tx and𝑦=𝑥 /𝑥^2+8

enclose a bounded region of positive area. When this is so, express the area of the region in terms of 𝑡t. 

I know that the bounds are 0 < T < 1/8 but I can't find the area, can someone help?

Answer 1

First lets determine where the two curves intersect.

Tx = x/(x²+8)

x = ±√(1/T - 8)

If we restrict ourselves to the first quadrant, then

we have just one intersection, call it, a = √(1/T - 8)

The area of interest is the difference between the definite integrals from 0 to a

∫(x/(x²+8)dx - ∫Txdx =

ln(x²+8)/2 - Tx²/2 from 0 to a =

(ln(a²+8) - ln(8) - Ta²)/2

(ln(1/T -8+8) - ln(8) - T(First lets determine where the two curves intersect.

Tx = x/(x²+8)

x = ±√(1/T - 8)

If we restrict ourselves to the first quadrant, then

we have just one intersection, call it, a = √(1/T - 8)

The area of interest is the difference between the definite integrals from 0 to a

∫(x/(x²+8)dx - ∫Txdx =

ln(x²+8)/2 - Tx²/2 from 0 to a =

(ln(a²+8) - ln(8) - Ta²)/2 =

(ln(1/T -8+8) - ln(8) - T(1/T -8))/2 =

(-ln(T) - ln(8) - 1 + 8T)/2

If we consider both the first and third quadrant, then by symmetry the area is double:

8T - ln(T) - ln(8) - 1