Welcome to a lesson on the definition
of a definite integral.
A definite integral is given using this notation here
where if a function f of x is continuous
on the closed interval from a to b,
then the def. integral of f of x integrated
with respect to x from a to b, is defined as the limit
as n approaches infinity of the sum of these products.
Now we should recognize the sum of these products
as the sums that we used to approximate the area
under a function and above the x-axis using rectangles.
So this is telling us that as the number of rectangles
bounded by the function of the x-axis approach infinity,
the sum of the area of those rectangles approaches the value
of the def. integral.
Let's look at an animation to better understand this.
Notice here's a non-negative function over
the closed interval from zero to eight
and we could use the area of these four rectangles
to approximate the area under the curve and above the x-axis
but as the number of rectangles increases,
our approximation of the area will get better and better.
So as the number of rectangles approaches infinity,
the area of these rectangles approach the value
of the def. integral.
Now notice in this example,
the width of each interval is the same,
but this doesn't have to be the case
and also notice that we're using
the right side of each interval to determine the height
of the rectangle, but of course,
we could also use the left side which would look like this.
In either case, as the number of rectangles approach
infinity the area of the rectangles approach the value
of the def. integral.
And as in this case, if the function is non-negative
then the def. integral of f of x integrated
with respect to x from a to b,
is equal to the area below the function and above the x-axis
over the closed interval from a to b.
Looking at the graph of y equals sine x below,
notice on the interval from zero to pi
the sine function is non-negative so the shaded area
would be equal to the integral of sine x
integrated with respect to x from zero to pi.
We don't know yet, but this area is exactly two square units
and therefore the value of this def. integral
would also be two.
Now if the function happens to be negative,
or below the x-axis on the closed interval,
then an integral of f of x from a to b would be
equal to the opposite of the area below the x-axis
on the closed interval from a to b.
So because of symmetry,
this orange area is also two square units
but because the area is below the x-axis,
this tells us the integral of sine x,
integrated with respect to x
on the interval from pi to two pi
would be equal to negative two.
And now if we have a function that is positive and negative,
for example here, the sine function on the closed interval
from zero to two pi
the value of the def. integral over the closed interval
from zero to two pi would be equal to the area
above the x-axis minus the area below the x-axis.
And because of the symmetry, this area is two square units
and so is this, which means that we wanted to find
the integral of sine x integrated with respect to x
from zero to two pi.
This would be equal to positive two, we integrate from
zero to pi plus we integrate from pi to two pi though,
that would be negative two giving us a value of zero.
Now we're concerned about the area under functions
for a variety of reasons and here's a simple example.
Let's say an empty take starts being filled with water
at a rate of four gallons per minute.
How much water is in the tank after 45 minutes?
Well a rate function would just be the constant function
r of t equals four.
If we graph this on the coordinate plane,
it would look like this graphed here in blue
because the tank is being filled for 45 minutes,
if we found the area under this function,
over the closed interval from zero to 45,
this would give us the amount of water in the tank
after 45 minutes, which would be the area under the function
over this closed interval.
And we could find this area using the area formula
for a rectangle, notice how the height would be four
and the width would be 45.
So we can say that the def. integral of four
integrated with respect to t from zero to 45 is equal to
this area which would be four times 45
which is equal to 180
which would be the number of gallons in the tank
after 45 minutes if it's being pumped
at a rate of four gallons per minute.
Let's take a look at another example.
Okay, we want to find the area under the function f of x
equals the absolute value of x minus two
and above the x-axis on the closed interval
from zero to five.
So we're trying to find the area of
these two shaded regions here.
And because we have two separate regions
and the function is non-negative,
we can represent this area as two separate def. integrals
given here below.
Notice how we first integrate from zero to two
and then we integrate from two to five.
And again we can evaluate these
because these regions are triangles
where the area of a triangle is equal to
1/2 the base times height.
So the first def. integral would be equal to the area of
this shaded region which would be 1/2 times the base of two
times the height of two plus,
then the integral from two to five would be equal to
this area here which would be 1/2 times the base,
which is two to five that's three units
times the height which is also three.
So the sum of these def. integrals is equal to
the sum of these products which would be
1/2 times two times two, that would be two,
plus, here we have 9/2.
Well two is equal to 4/2, 4/2 plus 9/2 would be 13/2.
So this is the value of the sum of these two def. integrals,
which is also the area of these two shaded regions.
Let's take a look at one more example.
We're given this graph
and we're told the areas of each region.
We want to use this to evaluate each
of the given def. integrals.
Let's go and label the areas.
This is eight square units.
A sub two is five square units.
A sub three is three square units,
and a sub four is also three square units.
So the value of the def. integral of f of x
integrated with respect to x from zero to three.
Notice how the function is non-negative over this interval,
is equal to this area here, which equals eight.
Next we have the integral of f of x from three to five,
you'll notice from three to five
the function is below the x-axis.
And since it's area is five square units
and the area is below the x-axis, this integral is
equal to negative five.
So now they want to find the integral of f of x
from zero to five, this would be equal to the sum of
the first two integrals
Well if we integrate from zero to three,
because the area is above the x-axis, that would be
positive eight plus,
when we integrate from three to five though,
we would get negative five.
Eight plus negative five is equal to positive three.
Next we have the integral of f of x from zero to seven.
So we integrate from zero to three, we'd have positive eight
plus we integrate from three to five,
that would be negative five,
when we integrate from five to seven,
notice how the area is above the x-axis,
so that would be positive three.
So eight plus negative five is three plus three
is positive six.
And then finally we have the integral of f of x
from three to ten.
So we're integrating from here to here.
So we integrate from three to five,
that would be negative five
because the area is below the x-axis
plus we integrate from five to seven,
the area is three, it's above the x-axis
so that's plus positive three
plus we integrate from seven to ten,
the area is below the x-axis,
and therefore it'd be negative three.
So the value of this def. integral would be equal to
negative five plus three plus negative three
which is equal to negative five.
Okay, hope you found this helpful.