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Find the area of the region bounded by π¦ equals π₯ cubed and π¦ equals π₯.
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Before we can begin to find the area of this region, we need to know what it looks like.
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And in order to see that, we need to produce a sketch of these two graphs.
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Thatβs π¦ equals π₯ cubed and π¦ equals π₯ on the same set of axes.
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We know what each of these graphs looks like individually.
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π¦ equals π₯ cubed is a simple cubic graph and π¦ equals π₯ is a straight line with a positive gradient.
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But we also need to know how these two graphs are positioned relative to one another.
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In order to see this, we need to find the π₯-coordinates of the points where the two graphs intersect.
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Now, each graph has been given in the form π¦ equals something.
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So we can set the two expressions for π¦ equal to one another to give an equation in π₯ only.
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Doing so gives π₯ cubed is equal to π₯.
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We can then subtract π₯ from each side of the equation, giving π₯ cubed minus π₯ equals zero.
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And to solve this equation, we need to take out a common factor of π₯ from each term.
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This gives π₯ multiplied by π₯ squared minus one is equal to zero.
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And in fact, we can go a step further because our second factor π₯ squared minus one is a difference of two squares.
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So it can be factored as π₯ minus one multiplied by π₯ plus one.
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Our equation has, therefore, become π₯ multiplied by π₯ minus one multiplied by π₯ plus one is equal to zero.
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And to solve, we need to take each factor in turn, set it equal to zero, and solve the resulting linear equation.
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We have first of all π₯ equals zero, then π₯ minus one equals zero, leading to π₯ equals one, and finally π₯ plus one equals zero, leading to π₯ equals negative one.
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And so, we find that these two graphs intersect at three places with π₯-coordinates of zero, one, and negative one.
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Now, itβs just worth mentioning that it is really important that we solve this equation by first factoring π₯ from our two terms.
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And we donβt instead divide the equation by π₯ to give π₯ squared minus one equals zero.
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If we had done this, then the only solutions would be π₯ equals positive and negative one.
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And we would have lost the solution π₯ equals zero.
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So make sure you bear that in mind.
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We donβt divide by π₯.
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We factor by π₯ if we can.
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And therefore, we make sure that we donβt lose any solutions.
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Now that we know the π₯-coordinates of the three points of intersection of these two graphs, we can sketch them on the same set of axes.
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We sketch π¦ equals π₯ cubed first of all.
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And then, we know that the graph π¦ equals π₯ needs to intersect this graph in three places: when π₯ equals negative one, when π₯ equals zero, and when π₯ equals one.
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We also know that π₯ cubed will be greater than π₯ for values of π₯ greater than one.
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So the graph of π¦ equals π₯ cubed will be above the graph of π¦ equals π₯ when π₯ is greater than one.
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So we add the line π¦ equals π₯ to our sketch.
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And now, we can see the area of the region that weβre looking for.
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Itβs the area of the full region enclosed by the two graphs.
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So the area now shaded in green.
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Now, notice that this area is made up of two distinct regions.
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And whatβs different is that in region one, the graph of π¦ equals π₯ cubed lies above the graph of π¦ equals π₯.
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Whereas in region two, the graph of π¦ equals π₯ lies above the graph of π¦ equals π₯ cubed.
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Letβs consider region two first of all.
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Now, we know that in order to find the area enclosed by a line and a curve, we can use integration.
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And what we need to do is subtract the equation of the lower curve from the equation of the upper curve.
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In this case, π¦ equals π₯ is the upper curve or upper line.
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And so, we have the integral of π₯ minus π₯ cubed with respect to π₯.
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The limits for this integral are the π₯-values where the two curves or the curve in the line intersect.
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So the limits will be zero and one.
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And we have that the area of region two is given by the definite integral from zero to one of π₯ minus π₯ cubed with respect to π₯.
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We can find the area of region one in a similar way.
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This time the limits are going to be negative one and zero.
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And here, the graph of π₯ cubed is above the graph of π₯.
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So we have the integral from negative one to zero of π₯ cubed minus π₯ with respect to π₯.
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Letβs perform integral two first of all.
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We recall that in order to integrate powers of π₯ not equal to negative one, we increase the power by one and then divide by the new power.
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So we have that the integral is equal to π₯ squared over two minus π₯ to the fourth power over four evaluated between zero and one.
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Now, this will all be equal to zero when π₯ is equal to zero.
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So substituting the limits, we just have one squared over two minus one to the fourth power over four, which is a half minus a quarter which simplifies to one-quarter.
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So we found the area of region two.
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And we can find the area of region one.
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By evaluating its integral this time we have π₯ to the fourth power over four minus π₯ squared over two evaluated between negative one and zero.
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That will all give zero when zero is substituted for π₯.
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So we have zero minus negative one to the fourth power over four minus negative one squared over two.
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Thatβs zero minus one-quarter minus a half or zero minus negative a quarter, which is equal to one-quarter.
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And so, we find that the area of region one is the same as the area of region two.
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In fact, we could have used the rotational symmetry of the graphs of both π¦ equals π₯ and π¦ equals π₯ cubed in order to see this.
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And so, we could have just found the area of one of these regions and then doubled it in order to find the total area.
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But if not, we can just add our two separate areas together: one-quarter plus one-quarter, which is equal to one-half.
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So using definite integration, we were able to find that the area of the region bounded by the curve π¦ equals π₯ cubed and the straight line π¦ equals π₯ is one-half.