Integration

Lesson

To complete our work on area, let's now turn our attention to finding the area between two curves.

In situations like this, it's useful to think of the area we're finding as the "triangle shape".

By graphing the curves $y=2x$`y`=2`x` and $y=2(x-2)^2$`y`=2(`x`−2)2, find the area bounded by the curves and the $x$`x`-axis.

By graphing both curves and shading the region of interest we get the following graph.

You can see the "triangle shape" here. To find the area, we need to find the area underneath the line up to the point of intersection of $x=1$`x`=1, and then the area underneath the quadratic function from the point of intersection.

Setting this up we have:

Now we go through the process of definite integration to find the total area.

When the area is wedged between two curves we can subtract the area bounded by the lower curve and the $x$`x`-axis away from the area bounded by the upper curve and the $x$`x`-axis.

By graphing the curves $y=6-2x$`y`=6−2`x` and y=$x(x-3)^2$`x`(`x`−3)2, find the area trapped between the two curves.

Graphing gives us the following two regions trapped between the curves.

Notice that on the left the cubic function is uppermost while on the right the linear functions is uppermost. So we'll need to set it up in two parts like this:

Because we need the area under the upper curve minus the area under the lower curve at all times, we need to create two calculations to accommodate where the curves switch positions at the intersection point.

And now to conduct the calculus!

How else could we have done this one?

Well, we had we realised that each shaded portion was the same in shape and size (a reflection of each other, or a rotation) we could have calculated one area and then multiplied by $2$2, saving us a lot of time!

If we have the functionality of a CAS calculator and this is a calculator-allowed question, then we can do the following:

Find the area enclosed between the lines $y=2x$`y`=2`x`, $y=\frac{1}{3}x$`y`=13`x` and $x=6$`x`=6.

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Find the area bounded between the two curves $y=\sqrt{x+5}$`y`=√`x`+5 and $y=-x-3$`y`=−`x`−3 and the $x$`x`-axis.

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Consider the functions $y=-x^2+8$`y`=−`x`2+8 and $y=-x+2$`y`=−`x`+2.

Graph the functions on the axes below.

Loading Graph...State the values of $x$

`x`at which the line and the parabola intersect.Write both values on the same line, separated by a comma.

Hence find the area enclosed between the line and the curve.

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

Apply integration methods in solving problems