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Find the area of the region enclosed by the curves π¦ is equal to 16 cos of π₯ and π¦ is equal to two sec squared of π₯ for π₯ between negative π by three and π by three.
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In this question, weβre asked to find the area of a region enclosed by two curves, π¦ is equal to 16 cos of π₯ and π¦ is equal to two sec squared π₯, and between two vertical lines, π₯ is equal to negative π by three and π₯ is equal to π by three.
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And whenever weβre asked to find an area of a region enclosed by curves, we should always sketch a diagram, because it can help us determine the easiest way to find this area.
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Before we start sketching our curves, we should note that our values of π₯ will range between negative π by three and π by three.
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So we only need to sketch these curves for these values of π₯.
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So letβs start by sketching π¦ is equal to 16 cos of π₯ between π₯ is negative π by three and π by three.
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We can do this by noticing this is just the graph π¦ is equal to cos of π₯ stretched by a factor of 16 in the vertical direction.
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So this is just a vertical stretch of the cosine function.
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Its graph looks something like the following.
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On the same axis, we need to sketch the curve π¦ is equal to two sec squared of π₯.
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And we can see that this will be a vertical stretch of factor two of the secant function.
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But remember, weβre sketching this on the same axis weβre sketching π¦ is equal to 16 cos of π₯.
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So we need to make sure we get our scales correct.
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One way of doing this is to consider a few points on our graph.
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First, we can find the π¦-intercept of π¦ is equal to 16 cos of π₯ by substituting π₯ is equal to zero into the function.
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cos of zero is one, so the π¦-intercept is 16.
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We can then find the π¦-intercept of our other curve.
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We substitute π₯ is equal to zero to get two sec squared of zero.
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The sec of zero is one, so the π¦-intercept is at two.
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So we can add this point onto our diagram.
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And now weβre almost ready to sketch our curve.
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We know the shape of this function.
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However, it will be useful to find the endpoints of this curve.
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Weβll do this by substituting π₯ is negative π by three and π₯ is π by three into the function.
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We can do both of these at the same time by noticing that the secant function is an even function.
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Therefore, two sec squared of negative π by three is equal to sec squared of π by three.
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Weβll rewrite this by noting the sec of π by three is one divided by the cos of π by three.
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So we get two divided by cos squared of π by three.
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Next, the cos of π by three is one-half.
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So we get two divided by one-half squared, which we can evaluate is equal to eight.
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However, at this point, we can notice something interesting.
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The cosine function is also an even function.
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And 16 cos of positive or negative π by three is equal to eight.
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So the π¦-coordinates of the endpoints of our curve π¦ is equal to 16 cos of π₯ is eight.
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So the endpoints of the two curves meet up, which give us a sketch which looks like the following.
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We can now shade in the region we need to determine the area of.
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We can now see that this region is bounded above by the curve π¦ is equal to 16 cos of π₯ and below by π¦ is equal to two sec squared of π₯.
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And itβs bounded between two vertical lines: π₯ is negative π by three and π₯ is equal to π by three.
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We can then recall if we have two integrable functions π and π and π of π₯ is greater than or equal to π of π₯ on a 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₯.
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And we can see this is true of the region we need to find the area of.
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π of π₯ is the upper function, 16 cos of π₯.
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π of π₯ is the lower function, two sec squared of π₯.
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π is the lower bound of π₯, negative π by three.
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And π is the upper bound of π₯, π by three.
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Finally, itβs also worth pointing out we know that our two functions are integrable because theyβre continuous on this interval.
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Therefore, substituting this information into the integral, we get the area of the shaded region is the integral from negative π by three to π by three of 16 cos of π₯ minus two sec squared of π₯ with respect to π₯.
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So we now need to evaluate this integral.
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We can do this term by term by recalling two integral results.
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First, we recall the integral of the cos of π₯ with respect to π₯ is the sin of π₯ plus the constant of integration πΆ.
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Second, we recall the integral of the sec squared of π₯ with respect to π₯ is the tan of π₯ plus the constant of integration πΆ.
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Of course, in this case, weβre working with definite integrals, so we donβt need to include the constants of integration.
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Therefore, 16 sin of π₯ is an antiderivative of 16 cos of π₯ and negative two tan of π₯ is an antiderivative of negative two sec squared of π₯.
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This gives us 16 sin of π₯ minus two tan of π₯ evaluated at the limits of integration π₯ is negative π by three and π₯ is π by three.
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We now need to find the difference of the antiderivative evaluated at the two limits of integration.
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And this is given by 16 sin of π by three minus two tan of π by three minus 16 sin of negative π by three minus two tan of negative π by three.
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And we could just evaluate this expression here.
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However, we can simplify this by noticing both sine and tangent are odd functions.
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Therefore, sin of negative π by three is negative one times the sin of π by three.
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Similarly, the tan of negative π by three is negative one multiplied by the tan of π by three.
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If we then distribute the negative over our parentheses, we can see that we get another term of positive 16π by three and another term of negative two tan of π by three.
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Therefore, this expression is equal to 32 sin of π by three minus four tan of π by three.
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And now we can evaluate this expression by first recalling the sin of π by three is root three over two.
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So 32 sin π by three is root three over two multiplied by 32.
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We can cancel the shared factor of two in the numerator and denominator to get 16 root three.
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Similarly, we can recall the tan of π by three is equal to root three.
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If we multiply both sides of this equation through by negative four, we get negative four tan of π by three is negative four root three.
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We can then substitute both of these values into our expression.
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Weβll first clear some space.
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Then, we substitute these expressions.
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We get 16 root three minus four root three, which we can calculate is 12 root three, which is our final answer.
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And itβs worth noting this represents an area, so we could say that this is 12 root three square units.
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Therefore, we were able to show the area of the region enclosed by the curves π¦ is equal to 16 cos of π₯ and π¦ is equal to two sec squared of π₯ for values of π₯ between negative π by three and π by three is 12 root three.