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If $R$ is the region bounded by the graphs of the functions $f(x)=\frac{x}{2}+5$ and $g(x)=x+\frac{1}{2}$ over the interval $\left[1,5\right],$ find the area of region $R.$
$12$ units ^{2}
In [link] , we defined the interval of interest as part of the problem statement. Quite often, though, we want to define our interval of interest based on where the graphs of the two functions intersect. This is illustrated in the following example.
If $R$ is the region bounded above by the graph of the function $f(x)=9-{\left(x\text{/}2\right)}^{2}$ and below by the graph of the function $g(x)=6-x,$ find the area of region $R.$
The region is depicted in the following figure.
We first need to compute where the graphs of the functions intersect. Setting $f(x)=g(x),$ we get
The graphs of the functions intersect when $x=6$ or $x=\mathrm{-2},$ so we want to integrate from $\mathrm{-2}$ to $6.$ Since $f(x)\ge g(x)$ for $\mathrm{-2}\le x\le 6,$ we obtain
The area of the region is $64\text{/}3$ units ^{2} .
If R is the region bounded above by the graph of the function $f(x)=x$ and below by the graph of the function $g(x)={x}^{4},$ find the area of region $R.$
$\frac{3}{10}$ unit ^{2}
So far, we have required $f(x)\ge g(x)$ over the entire interval of interest, but what if we want to look at regions bounded by the graphs of functions that cross one another? In that case, we modify the process we just developed by using the absolute value function.
Let $f(x)$ and $g(x)$ be continuous functions over an interval $\left[a,b\right].$ Let $R$ denote the region between the graphs of $f(x)$ and $g(x),$ and be bounded on the left and right by the lines $x=a$ and $x=b,$ respectively. Then, the area of $R$ is given by
In practice, applying this theorem requires us to break up the interval $\left[a,b\right]$ and evaluate several integrals, depending on which of the function values is greater over a given part of the interval. We study this process in the following example.
If R is the region between the graphs of the functions $f(x)=\text{sin}\phantom{\rule{0.2em}{0ex}}x$ and $g(x)=\text{cos}\phantom{\rule{0.2em}{0ex}}x$ over the interval $\left[0,\pi \right],$ find the area of region $R.$
The region is depicted in the following figure.
The graphs of the functions intersect at $x=\pi \text{/}4.$ For $x\in \left[0,\pi \text{/}4\right],$ $\text{cos}\phantom{\rule{0.2em}{0ex}}x\ge \text{sin}\phantom{\rule{0.2em}{0ex}}x,$ so
On the other hand, for $x\in \left[\pi \text{/}4,\pi \right],$ $\text{sin}\phantom{\rule{0.2em}{0ex}}x\ge \text{cos}\phantom{\rule{0.2em}{0ex}}x,$ so
Then
The area of the region is $2\sqrt{2}$ units ^{2} .
If R is the region between the graphs of the functions $f(x)=\text{sin}\phantom{\rule{0.2em}{0ex}}x$ and $g(x)=\text{cos}\phantom{\rule{0.2em}{0ex}}x$ over the interval $\left[\pi \text{/}2,2\pi \right],$ find the area of region $R.$
$2+2\sqrt{2}$ units ^{2}
Consider the region depicted in [link] . Find the area of $R.$
As with [link] , we need to divide the interval into two pieces. The graphs of the functions intersect at $x=1$ (set $f(x)=g(x)$ and solve for x ), so we evaluate two separate integrals: one over the interval $\left[0,1\right]$ and one over the interval $\left[1,2\right].$
Over the interval $\left[0,1\right],$ the region is bounded above by $f(x)={x}^{2}$ and below by the x -axis, so we have
Over the interval $\left[1,2\right],$ the region is bounded above by $g(x)=2-x$ and below by the $x\text{-axis,}$ so we have
Adding these areas together, we obtain
The area of the region is $5\text{/}6$ units ^{2} .
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