Welcome to an example
of the Second Fundamental
Theorem of Calculus.
The Second Fundamental Theorem of Calculus
is stated here below
where if f is continuous
on an open interval I
containing the constant a,
then for every x in the interval,
the derivative with respect to x
of the integral of f of t from a to x
is equal to f of x.
So, a couple things to notice here.
The lower limit of integration a must be
in an interval where f of t is continuous.
The upper limit of integration is x,
the same variable in
which we're integrating
with respect to.
And then to evaluate the
derivative of this integral,
we simply substitute x
for t in our function f.
The reason this can be helpful is
even if we're not able to
evaluate this integral here,
we can still find this
derivative using this theorem.
So in our example, we'll actually
first apply this theorem.
And then because we
can find this integral,
we'll actually find the integral
and then find the derivative
to verify the results.
So if f of x is equal to this integral,
we want to find f prime
of x and f prime of three.
So again, the first
thing we should recognize
is that zero is in an interval
where the function f of t
equals two t to the third is continuous.
Notice this function is
continuous everywhere.
And then second, the
upper limit of integration
is the variable x.
So if this integral is equal to f of x,
that means f prime of x
would be equal to the derivative
with respect to x of our integral
which means the derivative
of this integral
would just be equal to the function
where we substitute x for t.
So f prime of x is just
equal to two x to the third.
So that's the first part of this question,
f prime of x equals two x to the third.
And therefore, to find f prime of three,
we simply substitute three for x.
So that would be two
times three to the third.
Three to the third is equal to 27.
27 times two is equal to 54.
But again, let's also verify this
by finding this integral and then finding
the derivative of the integral.
So if f of x is equal to the integral
from zero to x of two t to the third dt,
if we want to evaluate this integral
we would first find the
antiderivative function
which would be two times t to the fourth
divided by four evaluated at x and zero.
Well, this would just be
one-half t to the fourth.
So we would first substitute x for t
and then substitute zero for t.
Substituting x for t would give us
one-half x to the fourth.
Substituting zero for t
would just give us zero.
So f of x is just equal to
one-half x to the fourth
which means f prime of x,
applying the power rule
of differentiation,
we would have one-half
times four x to the third
which is two x to the third.
So notice how doing this the long way,
f prime of x is still two x to the third.
I hope you found this helpful.
