Hi everyone.
Welcome back to integralcalc.com.
Today we’re going to be talking about how to find the area between two curves and the
functions we're given are y equals x plus 1 and y equals 9 minus x squared and we've
been asked to find the area between the two of them between x equals negative 1 and x
equals 2.
The formula that we're going to be using for this problem is a formula we always use for
area between curves problems and we're going to be integrating on the range a to b.
Our limits of integration a to b are going to be negative 1 to 2, which had already been
proivided for us, which is not always the case.
And then we're going to be integrating f of x minus g of x.
Now you have to decide which of your functions is going to be f of x and which is going to
be g of x.
f of x should be above g of x on the range negative 1 to 2.
So one way we could do that is by graphing both functions.
You can see that y equals 9 minus x squared is the purple graph here and y equals x plus
1 is the orange graph.
If we graph them and then we have here the line x equals negative 1 and then x equals
2, you can see that 9 minus x squared is above x plus 1 on the range negative 1 to 2.
So that gives us an indication that f of x is going to be 9 minus x squared and g of
x is going to be x plus 1.
The other way you could figure this out is plug in any number that's between negative
1 and 2 and plug them in to both functions and whichever one is higher, would be f of
x.
The lower one would be g of x.
So having determined that 9 minus x squared is f of x, we can go ahead and plug in both
of our functions to the integral here.
So now that we've got this set-up, we're going to start simplifying.
We'll end up with 9 minus x squared minus x and minus 1.
The negative sign carries over to both of these terms here.
When we combine constants, we'll end up with 8 minus x squared minus x.
And then we can start integrating.
When we take the integral of this first term here, 8, we'll get 8x.
Taking the integral of negative x squared, we add 1 to the exponent to get 3 and then
we divide the coefficient which is the 1 by the new exponent 3.
And same here with this x term, we add 1 to the exponent. 1 plus 1 gives us 2 and we divide
our coefficient 1 by our new exponent which is 2.
And then we're going to be evaluating between negative 1 and 2.
When we plug in our definite integral numbers here, we plug in 2 first because it's the
top number and then we always subtract here in the center and then plug in our bottom
number which is negative 1.
And when we start simplifying, we get 16 here on the first term.
2 cubed is 8 so we end up with a negatie 8/3 here on the second term.
2 squared is 4, 4 times 1/2 is 2 so we end up with a negative.
And then of course a negative 8 here for 8 times negative 1.
Negative 1 cubed is negative 1 so we end up with negative 1 times negative 1/3 gives us
a positive 1/3 and then negative 1 squared here is a positive 1 so we end up with a negative
1/2 on this last term.
And then when we distribute the negative sign here on the middle, we end up with a positive
8 because we have a negative times a negative here and negative 1/3 and a positive 1/2.
Now
the following are just basic simplification steps.
Now we're going to find the lowest common denominator which is 6.
Here, this 22, we'll multiply by 6/6 to get 132/6, we're multiplying 9/3 by 2/2 to get
18/6 and we're multiplying 1/2 by 3/3 to get 3/6.
So now that everything is over 6, we combine them.
We get 117 divided by 6 which if we divide both numerator and denominator by 3, we end
up with 39/2.
So that's our final answer.
39/2 is the area between the graph of 9 minus x squared and the line x plus 1.
So I hope that helped you, guys and I'll see you in the next problem.
Bye.