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.
the integral from 1 to 3 ___ dt= __
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.
So we have:
∫f(t) dt from 1 to 3 = ln(3^2) - ln(1^2)
= ln(9) - ln(1)
= ln(9)
Well, isn't this integral just a rollercoaster ride? Hang on tight because we're about to have some fun!
First, let's find f(t), the derivative of F(t):
F(t) = ln(t^2)
f(t) = F'(t) = (2t) / t^2 = 2/t
Now, let's integrate f(t) from 1 to 3 using the fundamental theorem of calculus. "Fundamental" sounds serious, but don't worry, we'll make it lighthearted!
∫[1 to 3] (2/t) dt
Now, let's evaluate this integral. Grab your calculator because things are about to get numerical!
∫[1 to 3] (2/t) dt = 2 * ln(t) evaluated from 1 to 3
Substituting the upper limit into the function:
= 2 * ln(3)
And now subtracting the function evaluated at the lower limit:
= 2 * ln(3) - 2 * ln(1)
Since ln(1) is equal to 0, that term disappears:
= 2 * ln(3)
So, the integral from 1 to 3 of f(t) dt is 2ln(3). Now, that's a result worth chuckling about!
To find the integral from a to b of f(t) dt, we use the fundamental theorem of calculus. According to the fundamental theorem of calculus, if F(t) is an antiderivative of f(t), then the definite integral of f(t) from a to b is given by F(b) - F(a).
In this case, F(t) = ln(t^2), and we need to find the antiderivative or derivative of F(t) to find f(t). Taking the derivative of F(t), we have:
f(t) = (ln(t^2))'
To find the derivative of ln(t^2), we can use the chain rule. The derivative of ln(u) is given by 1/u times the derivative of the inside function. So, applying the chain rule, we have:
f(t) = (ln(t^2))' = (1/(t^2)) * (2t) = 2/t
Now that we have f(t), we can find the integral from a to b.
The integral from 1 to 3 of f(t) dt is given by:
∫[1 to 3] (2/t) dt
To evaluate this integral using the fundamental theorem of calculus, we substitute the limits of integration (a = 1, b = 3) into the antiderivative F(t) = ln(t^2):
F(3) - F(1)
= ln(3^2) - ln(1^2)
= ln(9) - ln(1)
= ln(9)
Therefore, the integral from 1 to 3 of f(t) dt is ln(9).
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