for the function F(t)=ln(t^2), let f(t)=F'(t). Write the integral from a to b of f(t) dt and evaluate it with the fundamental theorem of calculus.

Question

the integral from 1 to 3 ___ dt= __

Answer 1

F(t) = ln t^2
= 2 lnt

f(t) = 2/t

so if you want ∫f(t) dt from 1 to 3
= ln t^2 from 1 to 3
= ln9 - ln1
= ln9 - 0
= ln 9

Answer 2

The fundamental theorem of calculus tells us that

b b
∫f(t) dt = F(t)│ = F(b) - F(a)
a a

which basically means that you can evaluate an integrand f(x) by using its antiderivative F(x). Specific to your problem,

b b b
∫f(t) dt = F(t)│ = ln(t^2)│ = ln(b^2)-ln(a^2)
a a a

Plug in 1 and 3 for your a and b.

Answer 3

So we have:

∫f(t) dt from 1 to 3 = ln(3^2) - ln(1^2)

= ln(9) - ln(1)

= ln(9)