### Video Transcript

Determine the area of the region bounded by the graphs of the functions π and π, where π of π₯ is equal to π₯ squared minus 11 and π of π₯ is equal to negative π₯ plus five squared plus six.

In this question, weβre asked to find the area of a region bounded by the graphs of two functions, the function π and the function π. And weβre given these two functions. We can see these are both quadratic functions. Weβre given π of π₯ is π₯ squared minus 11 and π of π₯ is equal to negative π₯ plus five squared plus six. And whenever we are asked to find the area of a region bounded by the graphs of functions, itβs always a good idea to sketch the region we need to find.

To do this, weβre going to need to sketch the graphs of the functions π¦ is equal to π of π₯ and π¦ is equal to π of π₯. Letβs start by sketching π¦ is equal to π of π₯ on the coordinate plane. To do this, we can start by factoring π₯ squared minus 11 by using a difference between two squares. This then gives us that the curve π¦ is equal to π of π₯ is π¦ is equal to π₯ minus root 11 multiplied by π₯ plus root 11. And this is now a quadratic equation given in factored form, so we can use this to sketch the quadratic.

First, we can look at the leading coefficient of the quadratic. We can see that the leading coefficient is positive. Therefore, our parabola will open upwards. Next, since weβve written our quadratic in factored form, we can see it will have two π₯- intercepts, one at root 11 and one at negative root 11. Finally, although itβs not necessary, we can substitute π₯ is equal to zero into our function to find the π¦-intercept. We would get the π¦-intercept is negative 11. This gives us the following sketch. Itβs also worth pointing out we couldβve sketched this diagram by noticing π of π₯ is a translation 11 units downwards of the curve π¦ is equal to π₯ squared.

We now also need to sketch π¦ is equal to π of π₯ on the same coordinate plane. This time, we can see that π of π₯ is a quadratic given in vertex form. So we can use this to find the coordinates of the vertex of the curve π¦ is equal to π of π₯. We see when π₯ is equal to negative five, the expression inside our parentheses is equal to zero, and this leaves us with six in the expression. So the coordinates of the vertex are negative five, six. And at this point, we can see it can be quite difficult to sketch the curve onto the same coordinate plane.

One way of determining where the vertex lies on this diagram is to substitute π₯ is equal to negative five into the function π of π₯. We get π evaluated at negative five is equal to negative five squared minus 11, which we can then evaluate as 25 minus 11, which is equal to 14. So when π₯ is equal to negative five, our function π outputs 14. However, π of π₯ is outputting six, so π of π₯ is lower than π of π₯ at this value of π₯. We can then add the coordinates of the vertex of the curve π¦ is equal to π of π₯ onto our diagram, where we know it needs to be below the curve π¦ is equal to π of π₯.

Weβre now almost ready to sketch the curve π¦ is equal to π of π₯. However, it can be useful to find the π¦-intercept of this curve first. We do this by substituting π₯ is equal to zero into the function π of π₯. We get negative one times zero plus five squared plus six. This gives us negative 25 plus six, which is equal to negative 19. And remember, the π¦-intercept of the curve π¦ is equal to π of π₯ was at negative 11. So the π¦-intercept of π¦ is equal to π of π₯ is lower than that of π¦ is equal to π of π₯. This then allows us to sketch the curve π¦ is equal to π of π₯, where we also note that the leading coefficient of π of π₯ is negative. So itβs a parabola which opens downwards.

We can now see the curves intersect at two distinct points, where the π₯-values are both negative. And we can also add the shaded region we need to find the area of to our diagram. Over this entire region, we can see that π of π₯ is greater than π of π₯, so we can use one of our rules for integration to determine this area.

We recall if we have two functions π of π₯ and π of π₯, where π of π₯ is greater than or equal to π of π₯ on the closed interval from π to π, then the area between the curves π¦ is equal to π of π₯ and π¦ is equal to π of π₯ and the vertical lines π₯ is equal to π and π₯ is equal to π is given by the definite integral from π to π of π of π₯ minus π of π₯ with respect to π₯. And itβs worth pointing out that this assumes that π of π₯ and π of π₯ are integrable on the closed interval from π to π. And we can see that this holds true for the area of the region in our diagram.

First, our values of π and π will be the π₯-coordinates of the two points of intersections between the two curves, where of course π is the larger of these two values. We can then see that π of π₯ is greater than or equal to π of π₯ on the closed interval from π to π. And the region is of course bounded by the two curves and the two vertical lines. Finally, π of π₯ and π of π₯ are both polynomials, so theyβre integrable on the entire set of real numbers. Therefore, to determine this area, we now need to find the two coordinates of the points of intersection between π¦ is equal to π of π₯ and π¦ is equal to π of π₯.

Weβll do this by setting π of π₯ equal to π of π₯ and solving for π₯, since we only need the π₯-coordinates anyway. We have π₯ squared minus 11 is equal to negative one times π₯ plus five squared plus six. Weβll now distribute the exponent over the parentheses. On the right-hand side of our equation, this gives us negative one times π₯ squared plus 10π₯ plus 25 plus six. We can then distribute the negative over the parentheses. And on the right-hand side of our equation, this gives us negative π₯ squared minus 10π₯ minus 25 plus six. We can now rearrange our equation by collecting like terms. Doing this, we get two π₯ squared plus 10π₯ plus eight is equal to zero. We can simplify this equation by dividing through by two. This gives us π₯ squared plus five π₯ plus four is equal to zero.

Now, we can solve this by factoring. We need two numbers which multiply to give four and add to give five. And we can see that this is one and four. So we get π₯ plus one multiplied by π₯ plus four is equal to zero. And finally, for a product to be equal to zero, one of the factors must be equal to zero. So either π₯ plus one is equal to zero or π₯ plus four is equal to zero. We can solve both of these equations to get π₯ is negative one or π₯ is negative four. These are the two π₯-intercepts of the point of intersection between the two curves. Remember, the lower one is the value of π, and the larger one is the value of π. So our value of π is negative four, and our value of π is negative one. We can now find the area of the region we need to determine.

By applying our integral rule, we have the area of the shaded region is the integral from negative four to negative one of negative one times π₯ plus five squared plus six minus π₯ squared minus 11 with respect to π₯. We could now start distributing and simplifying our integrand. However, this is not necessary. Weβve already found an expression for π of π₯ minus π of π₯ when we were solving for the π₯-coordinates of the points of intersection. So, in fact, we can multiply this by negative one to find an expression for π of π₯ minus π of π₯. We get that π of π₯ minus π of π₯ is equal to negative two π₯ squared minus 10π₯ minus eight.

And now, we can evaluate this integral by using the power rule for integration. We add one to the exponent of π₯ and then divide by this new exponent for each term. We get negative two π₯ cubed over three minus 10π₯ squared over two minus eight π₯ evaluated at the limits of integration, π₯ is equal to negative four and π₯ is equal to negative one. And we can simplify this slightly. We have 10 divided by two is equal to five.

Now, all thatβs left to do is evaluate our antiderivative at the limits of integration. Letβs start with the upper limit of integration π₯ is negative one. We get negative two times negative one cubed over three minus five times negative one squared minus eight times negative one. And if we evaluate this expression, we get 11 divided by three. Finally, we need to subtract the antiderivative evaluated at negative four. Evaluating the antiderivative at negative four, we get negative two times negative four cubed over three minus five times negative four squared minus eight times negative four. Then, we can note that the antiderivative evaluated at negative four simplifies to give us negative 16 over three.

Therefore, weβve shown the area of the shaded region is equal to 11 divided by three minus negative 16 over three, which we can evaluate is equal to nine. And since this represents an area, we will give this to units square units. Therefore, we were able to show the area of the region bounded by the graph of the functions π and π is nine square units.