Video: APCALC02AB-P1B-Q44-932121868072

What is the area between the curves 𝑦 = π‘₯Β³ βˆ’ π‘₯Β² βˆ’ π‘₯ + 2 and 𝑦 = 2π‘₯Β² + 5π‘₯ βˆ’ 6 from π‘₯ = βˆ’2 to π‘₯ = 4?

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Video Transcript

What is the area between the curves 𝑦 equals π‘₯ cubed minus π‘₯ squared minus π‘₯ plus two and 𝑦 equals two π‘₯ squared plus five π‘₯ minus six from π‘₯ equals negative two to π‘₯ equals four?

We’re going to answer this question using integration. But before we can perform our integration, we need to know what these two curves will look like relative to one another. Firstly, we can deduce the approximate shape of each curve. Our cubic curve has a positive leading coefficient. And therefore, it would have the shape drawn in orange. Our quadratic curve also has a positive leading coefficient. And therefore, it will have the shape drawn in pink.

However, we need to know the approximate positioning of these two curves relative to one another so that we can determine which is above the other in this interval we’ve been given. And to do that, we need to find their points of intersection. To find their points of intersection, we set the two expressions for 𝑦 equal to one another, giving a cubic equation. By collecting all terms on the left-hand side of the equation, we obtain π‘₯ cubed minus three π‘₯ squared minus six π‘₯ plus eight is equal to zero. And we now want to solve this cubic equation. So we need to see if it can be factored.

Now there’s actually a clue in the question because we’ve been asked to find the area between these curves from π‘₯ equals negative two to π‘₯ equals four. So it’s likely that these two values have been chosen because the two curves intersect at these π‘₯ values. If we define this cubic function to be β„Ž of π‘₯ and then we substitute both negative two and four into β„Ž of π‘₯, we find that both β„Ž of negative two and β„Ž of four are equal to zero. This tells us that the curves do indeed intersect when π‘₯ equals negative two and when π‘₯ equals four. And it also tells us that π‘₯ plus two and π‘₯ minus four are both factors of this cubic.

We could then perform polynomial division to find the third factor. Or we can observe that if we substitute one into our equation β„Ž of π‘₯, we also get zero. This tells us that π‘₯ minus one is the third factor of this cubic. And therefore, the two curves intersect in three places: when π‘₯ equals negative two, when π‘₯ equals one, and when π‘₯ equals four.

We can now combine this with the general shape of the two curves in order to sketch them on the same set of axes. We know that the cubic curve will be steeper than the quadratic curve for large values of π‘₯. So our two curves in relation to one another look like this, intersecting when π‘₯ equals negative two, π‘₯ equals one, and π‘₯ equals four.

The area we need to calculate is formed by two separate regions. And we need to be clear which curve is above the other in each of these regions. If we define the equation of the cubic curve to be 𝑓 of π‘₯ and the quadratic to be 𝑔 of π‘₯, then we see that, in the first region between π‘₯ equals negative two and π‘₯ equals one, 𝑓 of π‘₯ is greater than 𝑔 of π‘₯. The cubic curve is above the quadratic. Whereas in the second region between π‘₯ equals one and π‘₯ equals four, 𝑔 of π‘₯ is greater than 𝑓 of π‘₯. The pink curve, the quadratic, is above the orange curve, the cubic.

So the total area between these two curves then will be found by the integral from negative two to one of 𝑓 of π‘₯ minus 𝑔 of π‘₯. That’s the equation of the cubic minus the equation of the quadratic. Plus the integral from one to four of 𝑔 of π‘₯ minus 𝑓 of π‘₯. That’s the equation of the quadratic minus the equation of the cubic.

We can simplify each of these integrands by subtracting term by term. The integrand becomes π‘₯ cubed minus three π‘₯ squared minus six π‘₯ plus eight for our first integral. And we have the exact negative of this for our second integral, because we subtracted in the opposite order. We can now integrate term by term, recalling that, in order to integrate a power of π‘₯, where the power is not equal to negative one, we increase the power by one and then divide by the new power. We get π‘₯ to the fourth power over four minus three π‘₯ cubed over three minus six π‘₯ squared over two plus eight π‘₯, evaluated between negative two and one. And then the exact negative of this evaluated between one and four.

We recall that we do not need a constant of integration as this is a definite integral. And we can then simplify some of our coefficients. We can then substitute the limits for each integral. And we recall that one to the power of anything is just one. So each integral evaluated at one will just be equal to the sum of the coefficients. We obtain 17 over four minus negative 16 plus 16 minus negative 17 over four, which all simplifies to 81 over two. So we found that the area between the two given curves from π‘₯ equals negative two to π‘₯ equals four is 81 over two square units.

Now it’s just worth pointing out that because we do have access to a graphical calculator in this question, some of the earlier stages of working out could’ve been performed or at least checked on our calculators. For example, we could’ve used our calculators to solve the cubic equation and find the three π‘₯ values where the two curves intersect. Or we could’ve used our calculators to aid in sketching the two curves in relation to one another. It is important though that we’re able to perform these skills ourselves. But using our graphical calculators is a helpful check.

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