Area Between Curves Example 2

Uploaded by TheIntegralCALC on 24.02.2011

Hi everyone! Welcome back to Today we’re going to be doing another example
of finding the area between two curves. In this particular example we’re given two
functions, one is y equals x and the other is y equals x squared. The formula that we’re
going to be using to determine the area between these two curves is this formula here. Now,
the first thing that you’ll notice is that we have two functions here inside of our integral,
f of x and g of x. One thing more we’re going to need to determine is which of our
functions will be f of x and which one will be g of x, which will get to in a second.
The other thing you’ll notice about this formula is that we’re going to be integrating
on the range a to b. Now, sometimes we’re given our range or
limits of integration a and b. In this particular case, we’re not, so the first thing we have
to do is find our limits of integration a and b and the way that we’re going to do
that is by setting our two equations equal to one another. So we’re going to set x
equal to x squared and we’re going to solve this equation for x. So the way that we’ll
do that is subtract x from both sides to get this equation set equal to zero then we’ll
factor out an x on the right hand side and we’ll end up with x times x minus one. And
what that tells us is that, according to this term right here, x could be equal to zero
because, right, if x were equal to zero, we have zero out in front, it would… would
have zero multiplied by this whole term,  everything inside the parenthesis, and we’d get zero
equal zero which would satisfy our equation, or, according to this term, x could be equal
to one because we’d get one minus one here which would give us zero, we have this term
multiplied by zero, and that would also satisfy the equation, which means that x is equal
to either zero or one. So, zero and one are the points at which our
two functions, y equals x and y equals x squared, intersect one another. So, now that we’ve
determined that, the next thing we need to determine is, like I said before, which of
our functions will be f of x and which will be g of x. Now, when we’re finding the area
between curves, it’s really important to figure out which is which because f of x must
be the function that is greater than or literally higher than the other function on the range
at which we’re evaluating, so in this case, zero to one. So, if we look at a graph of
these two functions, we can see the purple line here is y equals x, the orange curve
is y equals x squared, and where they intersect one another, here and here, are the points
x equals zero and one which we just determined our… our… the… our two points of intersection
where our functions are equal to one another. So, you can see really easily that y equals
x is above or greater than y equals x squared on this range. So, two really easy ways to
determine which one is higher, either graph it and you’ll see right away or you could
pick any point in between our points of intersection, so, since our points of intersection are zero
and one, for example, you could pick one half, plug one half into both functions and whichever
one is higher is your higher equation and it’s going to be f of x. So, in our case,
clearly we’ve…we’ve seen that y equals x is greater than y equals x squared on the
range zero to one which means that, our integral, we’re going to plug in our limits of integration
where our functions intersect one another which we found to be zero and one so we’ll
plug in zero for a and one for b. We’ve determine that f of x is going to be equal
to x here because x is the function that’s greater on the range zero to one so we plug
in x here for f of x which means we’re plugging in x squared for g of x, so we plug in x squared
down here for g of x. Now, to simplify this equation for this integral, we just took away
the parenthesis here, we end up with x minus x squared and this is simple enough to integrate
right away. So what we’re going to be end up doing is
integrating and then evaluating on the range zero to one. So to integrate, we’ll take
it term by term. Looking at this x term here first, the exponent on this x term is one,
it’s implied, we don’t write it, but we have x to the one right here. We add one to
the exponent when we integrate so one plus one gives us two. We divide the coefficient
which is also an implied one by the new exponent two so we end up with one half x squared.
Same thing here, we add one to the exponent so two plus one gives us three and then we
divide our coefficient which is one by the new exponent three. So this is the result
after we integrate. And now, we’re going to evaluate on the range zero to one. So when
we do that, we always plug in the top number first, in our case, one, so you can see we’ve
plugged in one here for x. Then, we always subtract in the middle and then plug in the
bottom number which in our case is zero, so we plug that in to our equation. Now we’re
going to simplify. So our first term here, one half times one squared is one half, one
third times one cubed is one third so we have a minus one third, and both of these terms
are going to end up being zero’s so we end up with a minus zero plus zero because we
have a double negative here but it’s a… doesn’t really matter coz it’s just zero.
So that leaves us with one half minus one third. If we find the lowest common denominator
which is six and we modify here our fractions, we’ll end up with three sixth minus two
over sixth. And, when we simplify, you can see that our final answer is going to be one
sixth. So one sixth is the area between those two curves on the range zero to one and that’s
it, that’s our final answer. So I hope that helped you guys and I’ll
see you in the next video. Bye!