Welcome to a lesson on the
fundamental theorem of calculus.
The fundamental theorem of calculus
states that if a function f
is continuous on the closed interval from a to b
and big F is the antiderivative of little f
on the same closed interval,
then the def integral of f of x,
integrated with respect to x
from a to b,
is equal to big F of b
minus big F of a.
So this states that we can determine
the value of the def integral
by using the antiderivative function.
And here are some guidelines.
Number one,
provided we can find an antiderivative of f of x,
the integrand function,
we now have a much more efficient way to evaluate
a def integral.
Two, when applying the fundamental theorem of calculus,
the following notation is often used
where we first find the antiderivative function,
big F of x,
then this vertical bar means we evaluate this at b
and then subtract the value at a.
And three, it's not necessary to include
the constant of integration C in the antiderivative
because notice how if we did
and we find this difference,
the constant would simplify to zero,
leaving us with big F of b minus big F of a.
Let's look at some examples.
We want to evaluate the def integral.
And notice how the integrand function
graphed here on the right
is non-negative on the closed interval from one to three.
And, therefore, this def integral
is equal to the area below the function
bounded by the x-axis over this closed interval.
And notice how this region is a trapezoid,
so we could evaluate this
using the area formula for a trapezoid.
But let's use the fundamental theorem of calculus first.
So to apply the fundamental theorem of calculus,
we first want to find the antiderivative of two x plus one
with respect to x.
So we'd have two times,
this is x to the first,
so we'd have x to the second
divided by two plus,
the antiderivative of one with respect to x would be x.
We now need to evaluate the antiderivative at three
and then subtract the value of antiderivative
at one.
But before we do this, let's go ahead and simplify here.
This simplifies to just x squared plus x.
So now we evaluate the antiderivative first at three,
then at one,
and then find the difference.
So when x is three, we'd have three squared plus three.
When x is one, we'd have one squared plus one.
Simplifying here, we have three squared plus three,
that's 12,
minus,
here we have one squared plus one, that's two.
So the value of the def integral,
which is also this area, would be 10.
Now, for this first example,
let's verify these results
using the area formula for a trapezoid given here.
So let's call this b sub one, b sub two,
and this would be the height of the trapezoid.
So using our formula, we have the area is equal to 1/2
times the quantity b sub one plus b sub two,
that would be three plus seven.
And the height is two.
So we have 1/2 times 10 times two,
which is 10,
verifying our answer is correct.
Let's take a look at another example.
Again, notice how the integrand function is non-negative
over the closed interval from zero to three.
So this def integral is equal to this area here.
So applying the fundamental theorem of calculus,
we'll first find the antiderivative
of negative 1/2 x squared
plus two x plus two with respect to x.
So we'd have negative 1/2
times x to the third divided by three
plus two times,
this is x to the first,
so we'd have x to the second divided by two
plus the antiderivative of two with respect to x
would be two x.
So this is our antiderivative
which we now need to evaluate at three,
then zero,
then find the difference.
But, of course, we'll simplify this first.
So here we have negative 1/6 x to the third
plus, this would be x squared
plus two x.
So, again, we'll evaluate this at three
and then subtract the value at zero.
So when x is three, we have negative 1/6
times three to the third
plus three squared
plus two times three.
And when x is zero,
notice how each term would be zero.
So simplifying, we have,
this would be negative 1/6
times 27,
which is why this is 27 over one.
And then we have plus nine
and then plus six.
Notice how we have a common factor of three here.
There's two threes in six
and nine threes in 27.
So here we have a negative 9/2.
This would be plus 15
or plus 15 over one.
Common denominator is two.
So multiply 15 over one by two over two.
So we have negative 9/2 plus 30/2,
which is 21/2.
So this is the value of the def integral.
And because the integrand function is non-negative
on the closed interval from zero to three,
the area of the shaded region is 21/2 square units
or 10 1/2 square units.
Let's take a look at one more example.
We want to find the def integral of
the square root of x plus four
integrated with respect to x from one to four.
Once again, notice how the integrand function
is non-negative on this closed interval.
So the def integral is equal to this area here.
Before we find the antiderivative though,
we want to write the square root of x using rational exponents
where the exponent of x is one,
the index is two.
So we can write this as the integral of
x to the 1/2 plus four
integrated with respect to x from one to four.
And now we'll find the antiderivative function
with respect to x.
So for x to the 1/2,
we'd have x, the power of 1/2 plus one, that's 3/2,
divided by 3/2
plus, the antiderivative of four is four x.
But dividing by 3/2
is equivalent to multiplying by 2/3.
So write this as 2/3 x to the 3/2
plus four x.
And now we need to evaluate this first at four
and then subtract the value at x equals one.
So when x is four, we have
2/3
times four to the 3/2
plus four times four.
And then when x is one, we have 2/3
times one to the 3/2
plus four times one.
For four to the 3/2, that's equivalent to
the square root of four to the third,
which would be two to the third or eight.
So here we have 2/3 times eight, that would be 16/3
plus 16.
Then here we have 2/3 plus four,
or plus four over one, if we want.
Let's write 16 as 16 over one as well.
So [mumbles] common denominator of three.
Here as well.
We have 16/3 plus 48/3,
which is 64/3,
minus, here we have 2/3 plus 12/3,
so minus 14/3,
which is equal to 50/3.
So this is the value of our def integral,
which again, because the integrand is non-negative
on this closed interval,
is also the area of the shaded region.
Okay, I hope you found this helpful.