6.1 Areas between curves  (Page 2/4)

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 .

The region is depicted in the following figure.

The graphs of the functions intersect at $x=\pi \text{/}4.$ For $x\in \left[0,\pi \text{/}4\right],$ $\text{cos}x\ge \text{sin}x,$ so

On the other hand, for $x\in \left[\pi \text{/}4,\pi \right],$ $\text{sin}x\ge \text{cos}x,$ so

Then

The area of the region is $2\sqrt{2}$ units ² .

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 .