# Area Between Curves

1.Find the area of the region enclosed by the curves
$y=x^2$ and $y^2=x$
Include a graph of the region.
$y^2=x$ is actually the same as $y=x^{1/2}$ and this is the one "higher" on the graph, it will be the f(x). Which means $f(x)=x^{1/2}$. This means $y=x^2=g(x)$
In this situation we use argument one, because both functions are positive. So, the formula we will be using the formula below.
$$Area=\int^b_af(x)dx-\int^b_ag(x)dx$$
The next step is to find the antiderivaties of $f(x)$ and $g(x)$.
$$f(x)'=\frac{2}{3}x^{\frac{3}{2}}$$
$$g(x)'=\frac{1}{3}x^{3}$$
In order to find the a and b in the integral we must set $f(x)=g(x)$, so $x^2=x^{\frac{1}{2}}$. Now solve for $x$. $x^{\frac{1}{2}}(x^4-1)=0$ So, $x=0$ and $x=1$ and $a=0$ and $b=1$
Now we compute the integrals separately.
$f(x)$ Area would be equal to $$\frac{2}{3}(1)^{\frac{3}{2}}-\frac{2}{3}(0)^{\frac{3}{2}}=\frac{2}{3}$$
$g(x)$ Area would be equal to
$$\frac{1}{3}(1)^{3}-\frac{1}{3}(0)^3=\frac{1}{3}$$
The final step is taking $f(x)-g(x)$, so $\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$
So the area between the curves is $\frac{1}{3}$ units.
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2.Find the area of the region in the first quadrant enclosed by the graph of the function $f(x)=\frac{1}{x^2}$
and the line $10x+4y=21$.Include a graph of the region.
The first step is to graph the equations and decide what is the area we want.Now we need to find the points on the graph where the two functions intersect and we do this by setting them equal to each other. $$\frac{1}{x^2}=\frac{21-10x}{4}$$
When you solve this you'll find that $x=2$ and $x=\frac{1}{2}$. Now you can find the actual points of intersection by plugging the $x$-values back into one of the equations, and they are $(\frac{1}{2},4)$ and $(2,\frac{1}{4})$. The function that is higher is $10x+4y=21$. We will be using the following equation to solve for the area, and $f(x)$ will serve as the higher function. $$\int^b_af(x)-g(x)dx$$
Plugging in our funtions we will get: $$\int^2_{\frac{1}{2}}(\frac{21-10x}{4})-(\frac{1}{x^2})dx$$ Then it can simplify to: $$\frac{21}{4}x-\frac{5}{4}x+x^{-1} | ^2_{\frac{1}{2}}$$
Now we solve the equation by plugging in our $x$-values and taking their difference.
Plugging two into the equation will give you $\frac{92}{16}$ and when you put 0.5 into it you'll get $\frac{69}{16}$. The final step is taking the diiferece of the two and when you do that you get $\frac{23}{16}$. So the area under the curves is $\frac{23}{16}$ units.

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3. Find the area of the region in the first quadrant enclosed by the curves $y=cos(x)$ and $y=sin(x)$. Along with the y-axis.
The equation for the y-axis is $x=0$.
The first step is to find points of intersection.
We can do this by letting $x=0$ in $y=cos(x)$ and $y=sin(x)$
It is known that $cos(0)=1$ and that $sin(0)=0$.
$sin(x)=cos(x)$ is a little harder but we can get $\frac{sin(x)}{cos(x)}=1$ which goes to $tan(x)=1$ and that means $x=\frac{π}{4}$. That can be found by taking the inverse of the equation or graphing $tan(x)=1$ and reading the $x$-value. The $y$-value can be figured out by plugging $x=\frac{π}{4}$ into $y=cos(x)$. From that you get that $y=0.7071$. So all of the points of intersection are (0,1), (0,0), ($\frac{π}{4}$, 0.7071). This will be easier to solve if the Area equation is used in this way $\int^b_af(y)-g(y)dy$.
So, rewrite the functions in terms of $y$.
$y=cos(x)$ now becomes $x=sec(y)$
$y=sin(x)$ now becomes $x=csc(y)$
There will be two rectangles "drawn" in to the graph, so two integrals will have to intergrated.
$\int^{0.7071}_0csc(y)-0dy$$=$ $csc(y)+cot(y)|^{0.7071}_0$$=$$1.40086-0=1.40086$
$\int^1_{0.7071}sec(y)-0dy$=$sec(y)+tan(y)|^1_{0.7071}$=$1.557-1.6399$= $-0.0825$
Then add the two areas to get $1.318$ units as the final answer.