# How do you find the area bounded by y=4-x^2, the x and y axis, and x=1?

Dec 4, 2016

First picture what this region would look like by envisioning its graph (or just looking straight at it):

graph{4-x^2 [-9.54, 10.46, -3.92, 6.08]}

Thinking about how this is bounded from side to side, we see it's bounded by the $y$-axis and the line $x = 1$.

Since it's also bounded by the $x$-axis, we're looking for the positive area in that "slice" of the graph between $0 < = x < = 1$:

graph{(4-x^2)sqrt(x-x^2)/sqrt(x-x^2) [-2, 3, -1, 4.72]}

This are can be found through integrating the function from $x = 0$ to $x = 1$, or:

${\int}_{0}^{1} \left(4 - {x}^{2}\right) \mathrm{dx}$

Integrating (finding the antiderivative) and keeping the bounds gives:

$= {\left[4 x - {x}^{3} / 3\right]}_{0}^{1}$

$= \left[4 \left(1\right) - {1}^{3} / 3\right] - \left[4 \left(0\right) - {0}^{3} / 3\right]$

$= \left(4 - \frac{1}{3}\right) - \left(0 - 0\right)$

$= \frac{11}{3}$

The area under the specified curve with the specified bounds is $\frac{11}{3}$.