# Video: APCALC02AB-P1A-Q39-732171435106

What is the area enclosed by the curve π(π₯) = π₯β΄ β 9π₯Β² and the π₯-axis?

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### Video Transcript

What is the area enclosed by the curve π of π₯ equals π₯ to the fourth power minus nine π₯ squared and the π₯-axis?

To help us answer this question, letβs draw a sketch of the curve π of π₯. It is a quartic curve as the highest power of π₯ is four. And as we also have that term of negative nine π₯ squared, it will have this sort of shape. Weβll first find where the curve π of π₯ intersects the π₯-axis by solving π of π₯ equals zero. Thatβs π₯ to the power of four minus nine π₯ squared equals zero. We can first factor π₯ squared from the two terms on the left-hand side and then recognize that π₯ squared minus nine is the difference of two squares. So it factorizes as π₯ plus three multiplied by π₯ minus three. The fully factored form of π₯ to the fourth power of minus nine π₯ squared then is π₯ squared multiplied by π₯ plus three multiplied by π₯ minus three.

To find the π₯-coordinates of the points of intersection of the curve with the π₯-axis, we set each factor in turn equal to zero and then solve the resulting equations. We obtain π₯ equals zero as a repeated root because this is coming from the equation π₯ squared equals zero, π₯ equals negative three, and π₯ equals three. We can therefore sketch the curve π of π₯, intersecting the π₯-axis at negative three, a repeated root of zero, and positive three.

The area weβre looking to find then, the area enclosed by the curve and the π₯-axis, is the area shaded in orange. To find this area then, we need to integrate π of π₯. Thatβs π₯ to the fourth power minus nine π₯ squared with respect to π₯ between the limits of negative three and three. Now as both parts of this area are on the same side of the π₯-axis, we donβt need to worry about splitting this integral up into an interval between negative three and zero and an integral between zero and three. But as both parts of this area are below the π₯-axis, they will both be negative when we find the integral. We therefore say that the area is equal to negative the value we get when we perform this integral.

Now, weβre ready to integrate. We recall that to integrate 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. So the integral of π₯ to the fourth power with respect to π₯ is π₯ to the fifth power over five. In the same way, the integral of negative nine π₯ squared with respect to π₯ is negative nine π₯ cubed over three, which simplifies to negative three π₯ cubed. And we still have our limits of negative three and three. Remember we donβt need a constant of integration because this is a definite integral.

We can then substitute our limits and evaluate. We have three to the power of five over five minus three times three cubed. From this, weβre subtracting negative three to the power of five over five minus three times negative three cubed and thatβs all multiplied by negative one. We have the negative of 243 over five minus 81 minus negative 243 over five plus 81. 81 can be written as 405 over five. And simplifying gives negative 162 over five minus 162 over five all multiplied by negative one which simplifies to 324 over five. So we found that the area enclosed by the curve π of π₯ equals π₯ to the fourth power minus nine π₯ squared and the π₯-axis is 324 over five square units.

Itβs also worth pointing out that because this curve is symmetrical in the π¦-axis, we could have just found one-half of the area by integrating between zero and three say and then doubling it in order to find the total area. This wouldβve given us the same answer of 324 over five.