NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Area between sin/cos curve and linear function
Lesson

We can use integration to solve area problems. In this chapter, shapes are formed between the graphs of a linear function and a sine or cosine function, and techniques for calculating the areas of the shapes are explored.

 

Example 1

The following diagram shows a shape bounded by the graphs of $y=\sin x+1$y=sinx+1 and $y=-\frac{x}{2}+5$y=x2+5 and the vertical lines $x=\frac{\pi}{2}$x=π2 and $x=6$x=6.

The area of the enclosed shape must be the difference between the area under the graph of $y=-\frac{x}{2}+5$y=x2+5 and the graph of $y=\sin x+1$y=sinx+1 between $x=\frac{\pi}{2}$x=π2 and $x=6$x=6. The graphs do not go below the $x$x-axis so, we just do the integration

$\int_{\frac{\pi}{2}}^6\ \left(-\frac{x}{2}+5\right)\ \mathrm{d}x-\int_{\frac{\pi}{2}}^6\ \left(\sin x+1\right)\ \mathrm{d}x$6π2 (x2+5) dx6π2 (sinx+1) dx

or, slightly simpler,

$\int_{\frac{\pi}{2}}^6\ \left(-\frac{x}{2}+5-\left(\sin x+1\right)\right)\ \mathrm{d}x$6π2 (x2+5(sinx+1)) dx

which is,

$\int_{\frac{\pi}{2}}^6\ \left(-\frac{x}{2}-\sin x+4\right)\ \mathrm{d}x$6π2 (x2sinx+4) dx

Thus,

$\text{Area}$Area $=$= $\left[-\frac{x^2}{4}+\cos x+4x\right]_{\frac{\pi}{2}}^6$[x24+cosx+4x]6π2
  $=$= $-9+\cos(6)+24+\frac{\pi^2}{16}-0-2\pi$9+cos(6)+24+π21602π
  $\approx$ $10.29$10.29

 

 

Example 2

Suppose the right-hand boundary of the shape in Example 1 had been at the intersection of the line and the sine curve. To find the area of this new shape it will be necessary to find the $x$x-coordinate of the intersection point.

The intersection point is where $\sin x+1=-\frac{x}{2}+5$sinx+1=x2+5. That is, where $\sin x+\frac{x}{2}-4=0$sinx+x24=0.

This equation cannot be solved by ordinary algebraic methods but an approximation is obtainable using a graphing calculator or similar device implementing a numerical method. You should check that the intersection is at approximately $x=6.8785$x=6.8785.

As before, the area required is given by

$\text{Area}$Area  $=$= $\int_{\frac{\pi}{2}}^{6.8785}\ \left(-\frac{x}{2}-\sin x+4\right)\ \mathrm{d}x$6.8785π2 (x2sinx+4) dx
  $=$= $\left[-\frac{x^2}{4}+\cos x+4x\right]_{\frac{\pi}{2}}^{6.8785}$[x24+cosx+4x]6.8785π2
  $\approx$ $10.847$10.847

 

 

Example 3

Suppose now, we have a shape as in the following diagram. It is bounded by the line $y=1-\frac{x}{2}$y=1x2, the curve $y=\sin x+1$y=sinx+1 and the vertical line $x=\frac{3\pi}{2}$x=3π2. We need the enclosed area between $x=0$x=0 and $x=\frac{3\pi}{2}$x=3π2.

We might imagine that there is a complication because part of the shape is below the $x$x-axis, which will result in a negative area in the integration process. In fact, because we are finding the difference between a positive and a negative area, we automatically add the area below the axis, as required.

Another way to see this is to add a large enough number to each function to bring the entire shape above the $x$x-axis. This has been done in the following diagram. We added $2$2 to each function. This translates the shape vertically but does not change its area.

The calculation can now proceed as before.

If we had added any number $a$a to the functions, the integral would have been

$\int_0^{\frac{2\pi}{2}}\ \left(\sin x+1+a-\left(a+1-\frac{x}{2}\right)\right)\ \mathrm{d}x$2π20 (sinx+1+a(a+1x2)) dx

But this is the same as

$\int_0^{\frac{2\pi}{2}}\ \left(\sin x+1-\left(1-\frac{x}{2}\right)\right)\ \mathrm{d}x$2π20 (sinx+1(1x2)) dx

which is what we would have worked with in the first place.

 

Outcomes

M8-11

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

91579

Apply integration methods in solving problems

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