# Let R be the shaded region in the first quadrant enclosed by the y-axis and the graphs of y=4-x^2 and y=1+2sinx, how do you find the area?

Feb 21, 2016

$R = 1.764$ units squared

#### Explanation:

In area problems, the first thing to do is graph the functions in question so you have a visual sense of the problem. You can check out the functions in this example graphed here.

From the graph, we see that the region $R$ is the curvy-looking triangle. To find the area of $R$, we subtract the area of $1 + \sin x$ from the area of $4 - {x}^{2}$. The limits of integration are from the $y$-axis $\left(x = 0\right)$ to the point of intersection $\left(x = 1.102\right)$.

Putting this in math terms, we have:
$R = \left({\int}_{0}^{1.102} 4 - {x}^{2} \mathrm{dx}\right) - \left({\int}_{0}^{1.102} 1 + 2 \sin x \mathrm{dx}\right)$

The properties of integrals say we can split these two into mini-integrals like so:
$R = \left({\int}_{0}^{1.102} 4 \mathrm{dx} - {\int}_{0}^{1.102} {x}^{2} \mathrm{dx}\right) - \left({\int}_{0}^{1.102} 1 \mathrm{dx} + 2 {\int}_{0}^{1.102} \sin x \mathrm{dx}\right)$

Now we perform the integration:
$R = \left(4 {\left[x\right]}_{0}^{1.102} - {\left[{x}^{3} / 3\right]}_{0}^{1.102}\right) - \left({\left[x\right]}_{0}^{1.102} - 2 {\left[\cos x\right]}_{0}^{1.102}\right)$

And then evaluate:
$R = \left(4.408 - 0.446\right) - \left(1.102 - 2 \left(\cos 1.102 - \cos 0\right)\right)$
$R = 3.962 - \left(1.102 - 2 \left(0.452 - 1\right)\right)$
$R = 3.962 - \left(1.102 + 1.096\right)$
$R = 3.962 - 2.198$
$R = 1.764$ units squared