Tutorial: Area between two curves: Game version

Resources

Note To follow this tutorial, you should know how to use the definite integral to calculate the area between the graph of a function and the $x$-axis.

Fundamentals

In this tutorial we see how to calculate the area between the graphs of two functions $f$ and $g$ over an interval as shown:

Notice that, in the example shown, the curves do not cross in the interval $[a, b]$. We focus on this case first.

Area between two non-crossing graphs

In the situation illustrated above, where the graph of $f$ never dips below the graph of $g$ when $x$ is in the interval $[a, b]$ (so that $f(x) \geq g(x)$ for $x$ in $[a, b]$), the area of the region between the graphs of $f$ and $g$ and between $x = a$ and $x = b$ is

Integral of (Top − Bottom)

Caution If the graphs of $f$ and $g$ cross in the interval $[a, b]$, the above formula does not hold (see below).

Example

The graph of $f(x) = x+1$ never dips below the graph of $g(x) = x^2-1$ when $x$ is in the interval $[-1,1].$

%Therefore,

%Area \t $= \int_{a}^{b} [f(x) - g(x)]\.dx$ \\ \t $= \int_{-1}^{1} [(x+1) - (x^2-1)]\.dx$ \\ \t $= \int_{-1}^{1} (-x^2 + x + 2)\.dx$ \\ \t $= \Bigl\[-\frac{x^3}{3} + \frac{x^2}{2} + 2x\Bigr\]_{-1}^{1}$ \\ \t $= \Bigl(-\frac{1}{3}+\frac{1}{2}+2\Bigr) - \Bigl(\frac{1}{3}+\frac{1}{2}-2\Bigr) = \frac{4}{3}.$

Find the area enclosed by $y = %10, \ y = %11, \ x = %14$ and $x = %15.$

Area

%Q What about curves that cross
%A That is our next order of business:

Area between two crossing graphs

Suppose the curves cross at one or more points as illustrated here:

Then, notice that they do not cross inside any of the intermediate intervals $[a,c], [c,d]$ %and $[d,b],$ so we can calculate the areas between successive crossing points and add them:

Find the $x$-coordinates of all points where the curves cross. (Set $f(x) = g(x)$ and solve for $x.$).
Then, individually calculate the area between the curves for $x$ between successive crossing points as before.
Finally, add up the individual areas so obtained.

$\int_{a}^{c} [f(x) - g(x)]\.dx$

$\int_{c}^{d} [g(x) - f(x)]\.dx$

$ \int_{b}^{b} [f(x) - g(x)]\.dx .\qquad $

Integral of (Top − Bottom)

Note that, in each subinterval, we must decide which is the top function.

Example

As before, let $f(x) = x+1$ and $g(x) = x^2-1.$ This time, find the total area enclosed by the two curves and the vertical lines $x=0$ and $x=3.$

To find the $x$-coordinates of all points where the curves cross, we set $f(x) = g(x)$ and solve for $x:$

$x+1 = x^2-1 \quad $ \t $\Rightarrow \quad x^2-x-2 = 0$ \\ \t $\Rightarrow \quad (x+1)(x-2) = 0$ \\ \t $\Rightarrow x = -1$ %or $2$

Of these, only $x = 2$ lies strictly inside the interval $[0, 3]$ we are interested in. %Therefore,

%Area \t $= \int_{0}^{2} [(x+1) - (x^2-1)]\.dx$ $+$ $\int_{2}^{3} [(x^2-1) - (x+1)]\.dx$ \\ \t $= \int_{0}^{2} (-x^2 + x + 2)\.dx + \int_{2}^{3} (x^2 - x - 2)\.dx$ \\ \t $= \Bigl\[-\frac{x^3}{3} + \frac{x^2}{2} + 2x\Bigr\]_{0}^{2} + \Bigl\[\frac{x^3}{3} - \frac{x^2}{2} - 2x\Bigr\]_{2}^{3}$ \\ \t $= \Bigl(-\frac{8}{3}+2+4\Bigr) - 0 + \Bigl(9 - \frac{9}{2} - 6\Bigr) - \Bigl(\frac{8}{3}-2 -4\Bigr) = \frac{31}{6}.$

We will find the area enclosed by $y = %20, \ y = %21, \ x = %24$ and $x = %25.$

First, determine whether the graphs cross strictly inside the interval $[%24,%25]$. If they do, enter the values of $x$ where they cross (excluding the endpoints of the interval) separated by commas if there are more than one.

To see the detailed calculation and complete the tutorial you must first answer the above question correctly.

Now try the exercises in %4, some the %8, or move ahead to the next tutorial by pressing "Next tutorial" on the sidebar.

Left-most area

Right-most area