BAM! Mr. Tarrou. In this video we are going to look at four examples of problems that
deal with definite integrals to find areas bound by a function and the x axis. However
these four examples are going to be like these two up here where we have... or we are doing
the definite integral from -2 to 3 of 6 dx and the definite integral from 0 to 5 of 2x
dx. Now with this constant and this single degree function here, we might not need to
go through that limit process that we have been doing in a lot of these videos to find
this area. And indeed let's not forget that we are trying to find the definite integral
of a function. So when you see the integral of 6 dx, you are integrating f(x) is equal
to 6. And when you are integrating 2x dx, you are really taking the integral of f(x)
is equal 2x. Now this is just fancy notation for y and we are looking at a function of
just y=6. Well, the function or the graph of y=6 is a horizontal line. So let's see
what that looks like as a picture. Well we can see that when we graph y=6, here is our
horizontal line, and when we mark off the closed interval from -2 to 3... which we finding
that definite integral bound by those values... we can see that all we doing is trying to
find the area of a rectangle. So instead of going through that whole limit process, just
recognize that since we are taking the definite integral of a constant all we are really finding
is the area of a rectangle. That is just the area formula of length times width. So our
area is going to be equal to the width of this rectangle, and the distance between -2
and 3 is equal to 5. The height of the rectangle of course we have since we are bound between
the function and the x axis, we have the height of six. So our area is equal to 30 square
units. Now over here we have f(x)=2x. So again let me step out, make you a nice neat picture,
and see what shape... what geometric shape we are getting from this single degree function
between the closed interval of 0 to 5. So here we have our picture drawn. We have that
y=2x which of course is a line with a line with a slope of 2 and a y intercept of zero.
So when I graph that line and I mark off the closed interval between 0 and 5 we can see
that this definite integral of 2x dx between 0 and 5 is just finding the area of a triangle.
I don't need to go through the limit process or Riemann sum to figure out that area. I
can just do my old school Geometry formula which is area is equal one-half base times
height. It is going to be 1/2... our base of 0 to 5 is 5... and a height of 10. Half
of 10 is 5, and 5 times 5 is equal to of course 25 square units. I have two more examples.
One is going to develop into the shape of a trapezoid. Another one into the shape of
a semi-circle. And you know I just want to make you think about what it is you are taking
the definite integral of. If it is a constant, if it is a single degree, the circles are
a little bit trickier to recognize. But if you sketch the area of which you are trying
to find the definite integral of, find the area of, you may very well find that you don't
need to do quite so much work as we have been doing in the last few videos. Now a lot of
those examples were probably not.. you know.. able to do it with this kind of simplicity.
But when you have the opportunity please take advantage of it. Alright, last two examples.
Here we have the definite integral of -x+7 dx from 1 to 4, the closed interval from 1
to 4. Let me step off again, give you a nice neat picture, and recognize that shape and
apply the correct Geometric formula because this once again will not need to go through
that long limit process. Or you might want to just pause the video real quick and try
it yourself before I give you that answer. Ok. So we have now.. you can see that when
we have this closed interval between 1 and 4 and how it bound by the function and the
x axis, that we have the shape of a trapezoid. Now the area of a trapezoid is equal to, incase
you have forgotten, 1/2 times the height times b sub 1 plus b sub 2. So now the bases on
a trapezoid are the two sides that are parallel. Normally we see those as horizontal, so you
might want to think of this picture being rotated 90 degrees in either direction. Or
just recognize that these two parallel sides are going to act the bases of your trapezoid.
And that comes out to be 1/2... the height, well from 1 to 4 if you were to lay it down
on its parallel sides... 4-1 is equal to 3... so our height is 3... I don't know why I am
writing it like that. And the bases, well let's see here. We go from, at this x value
of 1, we have a y value of zero and we go up to 6. So that is a length of 6, now plus
this height here. When we take the x value of 4 and we plug it into our function -4+7
is equal to 3. So that height is equal to 3. This is 1/2 times 3 times... 6+3 which
is equal to 9. And that is 1/2 of, 3 times is 27, so of course we have 27/2 units squared
for this definite integral which is representing this area. Over here we have the definite
integral of the square root of 9-x^2 dx from the boundary of -3 to 3. So this might look
like a question that you need to go through the limit process, or the definition of area...
the Riemann sum... Whatever you teacher is calling it. Definite of Area is just a specialized
case or a specific case of Riemann Sum. If we take this out again remembering that we
are taking the integral of a function. We have f(x) is equal to the square root of 9-x^2.
Ok, well if you like... If you don't recognize what that shape is I am going to do two things
at once. I am going to square both sides to get rid of the radical, the square root symbol,
and I am going to also write f(x) as y. So we are going to take this function and we
are going to square both sides of it. And we get y squared, and the power of two of
course and the square root are going to cancel out. Those are inverse functions. We have
9-x^2. And we have an equation now with two squared terms. The x is squared and the y
is squared. That means it could be a circle which is a specialized type of ellipse, or
it could be a hyperbola. Hopefully it is nothing that fancy because we won't be able to just
take a basic formula from Geometry to solve it. But if we do take x squared and we add
it to both sides, we get x^2+y^2 is equal to 9. So this is... we have two squared terms,
we have coefficients which are equal and the denominators are also equal if you are tying
to pull back some concepts from learning your conic sections from Algebra 2 Honors or PreCalculus
or somewhere in there... and we have a 9 at the end. So we have a center... We have a
center with a center which is equal to (0,0) and whose radius is equal to the square root
of 9 which is going to be equal to 3. Now I did say that we have a circle. But, I am
not going to draw an entire circle when I put this graph up here in a second. The reason
why is I have manipulated the equation a little bit just to allow you to see what the actual
shape of the function is, or the graph is. But the original equation was f(x), or y,
is equal to the square root of 9-x^2. Now the author of the text book and me... I made
up the question... but as you are looking through your textbook, when you see that square
root in your textbook and there is no sign in front of it, there are not expecting you
to... again... to use both the positive and negative answer. They are only expecting you
to use the positive answer... right. Like the square root of 16 is 4 and -4. But if
the book says square root of 16 period, I guess it could be f(x) is the square root
of 16... They only want that, I think it is called the principal square root, the positive
square root. If they wanted you to use the negative answer, they would write negative
square root of blank or they would write plus/minus if they wanted you to use both. Now that is
different than if you were to start here and solve this for y. If you were to start here
and put your own square root into the function, then you would have to include both the positive
and negative answer. But when your textbook gives you the square root symbol only, or
any even root, with no sign in front we are only using the principal square root or the
positive square root. Alright , so let me get that picture drawn for you. Ok, so I have
got my semi-circle drawn. And I don't need vertical lines defining the closed interval
in which I am trying to find the definite integral because the function ends.... or
well... if you read left to right sort of begins at -3 and ends at 3. [x-intercepts]
So we don't need those vertical lines to define that closed interval. Indeed if it were we
probably would have to go through the limit process or the Riemann Sum process to find
the area.. we are going to find that definite integral representing that area. However we
don't. So we just find the area of a circle which is equal to pi r squared, now again
I don't want an entire circle... I just want a semi-circle or half of that, so I am going
to divide it by two. So our answer comes out to be the area is equal to the radius which
again is equal to 3... so we have pi times 3 squared over 2. And 3 squared is equal to
9, so we have 9pi/2. I am putting the square unit here because again when we are using
the definite integral to find that area bound by the function and the x axis. So since we
are identifying area, then it is square units for the unit of measure. I am Mr. Tarrou.
That is the end of my video for finding areas of common Geometric regions when you are doing
a definite integral. So... time for you to go do your homework. BAM!