### Video Transcript

What is the area, in square units, of the region enclosed by the graph π¦ equals negative two π₯ squared plus six and the line π¦ equals negative π₯ plus three.

As weβve been asked to find the area of a region enclosed by the graphs of these two lines, weβre going to use integration. But before we can do so, we need to know what these two graphs look like in relation to one another. π¦ equals negative two π₯ squared plus six is a quadratic with a negative leading coefficient. And so its graph will be a parabola with the shape shown. π¦ equals negative π₯ plus three is a straight line. And as the coefficient of π₯ is negative, it will have a negative slope. It will slope downwards from left to right. So we know the approximate shape of each of these graphs. But we also need to know where they intersect. Their equations have each been given in the form π¦ equal. So we can set the two expressions for π¦ equal to one another to give an equation in π₯ only. We have negative two π₯ squared plus six equals negative π₯ plus three.

We can group all the terms on the right-hand side by adding two π₯ squared and subtracting six. To give zero equals two π₯ squared minus π₯ minus three. And we now have a quadratic equation in π₯. We can solve this quadratic by factoring. We can either use the formal method of factoring by grouping. Or we can use trial and error. And we find that our quadratic factorises as two π₯ minus three multiplied by π₯ plus one. In order to find the points of intersection of these two graphs then, we need to take each bracket in turn and set it equal to zero. And then solve the resulting equation. Two π₯ minus three equals zero leads to π₯ equals three over two. And π₯ plus one equals zero leads to π₯ equals negative one. So these two lines intersect in two places. One with an π₯-coordinate of negative one. And the other with an π₯-coordinate of three over two.

We now have enough information to be able to sketch the two graphs. The quadratic, first of all, which is symmetrical in the π¦-axis and has a π¦-intercept of six. And then the straight line with a negative gradient and a π¦-intercept of three. We can clearly see the two points of intersection with the π₯-coordinates weβve already found of negative one and three over two. The area that weβre looking for then is this area here, enclosed by the two graphs. We find the area by evaluating a definite integral, with a lower limit of negative one and an upper limit of three over two. We could find this area by evaluating two separate integrals. As the quadratic curve is above the straight line for π₯-values between negative one and three over two. We could find the area below the pink curve and then subtract the area below the orange straight line.

However, itβs more efficient to perform a single integral. We subtract the equation of the straight line, which is the lower of the two graphs, from the equation of the quadratic curve. Giving the integral from negative one to three over two of negative two π₯ squared plus six minus negative π₯ plus three with respect to π₯. Simplifying the integrand gives the integral from negative one to three over two of negative two π₯ squared plus π₯ plus three with respect to π₯. And now we can integrate. We recall that, to integrate powers of π₯ not equal to negative one, we increase the power by one and then divide by the new power. So we have negative two π₯ cubed over 3 plus π₯ squared over two plus three π₯, evaluated between negative one and three over two. Thereβs no need for constant of integration as weβre performing a definite integral.

We can then substitute in our limits, giving negative two-thirds multiplied by three over two cubed. Plus a half multiplied by three over two squared. Plus three multiplied by three over two. Minus negative two multiplied by negative one cubed over three. Plus negative one squared over two plus three multiplied by negative one. Evaluating on a calculator gives 27 over eight minus negative 11 over six, which is equal to 125 over 24.

So by finding the π₯-coordinates of the points of intersection of these two graphs, we were able to determine the limits for our definite integral. By sketching the two graphs on the same set of axes, we were able to determine that the quadratic curve was above the straight line throughout this interval. We then found the area of the region enclosed by these two graphs by subtracting the equation of the straight line from the equation of the quadratic. And integrating between our limits of negative one and three over two. Giving an area of 125 over 24 square units.