Area Between Curves

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Horizontal Slicing

We can also compute areas by slicing horizontally instead of vertically. In this case, we'll denote by $W(y)$ the width of a horizontal strip. The area of each slice is roughly $W(y)\cdot\Delta y$, and the total area becomes $\int_c^d W(y)\,dy$, where $[c,d]$ is the $y$-interval that includes the total area to be computed. This means the horizontal strips start at the lowest $y$-value $c$, and end at the largest $y$-value $d$ -- these are not $x$-values here.

Let's look at an example. We have two curves below, $f$ and $g$, which are written in terms of $y$. As in the case of vertical strips, we need to know which is the larger. That translates to the curve further to the right (bigger $x$-values) being the larger.

In this example, $W(y)=f(y)-g(y)$. We compute our area:$$\int_c^d W(y)\,dy=\int_{-1}^1\left((f(y)-g(y)\right)\,dy=\int_{-1}^1(1-y^2-(y^3-y))\,dy=\int_{-1}^1(1-y^2-y^3+y)\,dy.$$ DO: Before looking ahead, evaluate this integral. Check your antiderivative!

$$\int_{-1}^1(1-y^2-y^3+y)\,dy=\left(y-\tfrac 1 3 y^3-\tfrac 1 4 y^4+\tfrac 1 2 y^2\right)\left|\begin{array}{c} ^1 \\ _{-1} \end{array}\right .=1-\tfrac 1 3-\tfrac 1 4+\tfrac 1 2 -(-1+\tfrac 1 3-\tfrac 1 4+\tfrac 1 2)=2-\tfrac 2 3 = \tfrac 4 3$$

In the example above, the two curves were naturally written as functions of $y$, since they are not functions of $x$ (why not?). However, we often are given functions of $x$, and we have to find the inverse function to get a function of $y$. You will see this in the video. What we do when we are given $y=f(x)$ is we find $f^{-1}$, since $x=f^{-1}(y)$.