# Submission Questions
## Advanced level

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1. Find the area of the region enclosed by the curves: $y=x^2$ and $x=y^2$
* First we must change $x=y^2$ to $y=\sqrt{x}$
* Next we set the two eqautions equal to each other in order to find the integrals we want to compute from:
$\sqrt{x}=x^2$

$(\sqrt{x})^2=(x^2)^2$

$x=x^4$ which then gives the x values of $+1$ and $-1$.
* Then we set up the integral:

$\int_0^5\sqrt{x}-x^2dx$ which can also be written as:$\int_0^5 x^\frac{1}{2} -x^2 dx$
* Let's now take the antiderivative of this integral:
$\frac{2}{3} x^\frac{3}{2} -\frac{1}{3} x^3$ from 0 to 1

$((\frac{2}{3} (1)^\frac{3}{2} -\frac{1}{3} (1)^3))+((\frac{2}{3} (0)^\frac{3}{2} -\frac{1}{3} (0)^3))=\frac{1}{3}$
* Final answer=$\frac{1}{3}$

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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 $10x+4y=21$.
* First we need to change $10x+4y=21$ to $y=\frac{21-x}{4}$
* Then we set both equations equal to each other:
$\frac{1}{x^2} =\frac{21-10x}{4}$ Simplifying this equation we get:

$4=x^2 (21-10x)$ $\rightarrow$ $21x^2 -10x^3 -4=0$ $\rightarrow$ $-1(10x^3 -21x^2 +4)=0$ $\rightarrow$ $(-1)(2x-1)(5x^3 -8x-4)=0$ $\rightarrow$ $-(2x-1)(5x+2)(x-2)=0$ $\rightarrow$ $(-2x+1)(5x+2)(x-2)=0$

So our $x$ values are $\frac{1}{2},\frac{-2}{5},2$

$\int_\frac{1}{2}^2 \frac{21-10x}{4}dx -\frac{1}{x^2}dx$

$\int_\frac{1}{2}^2 \frac{x^2 (21-10x)-4}{4x^2} dx$ $\rightarrow$ $\frac{1}{4} \int_\frac{1}{2}^2 \frac{(21-10x)x^2 -4}{x^2} dx$ $\rightarrow$ $\frac{1}{4} \int_\frac{1}{2}^2 ((21-10x)x^2 -4)*x^{-2} dx$ $\rightarrow$ $\frac{1}{4} \int_\frac{1}{2}^2 (21x^2 -10x^3 )-4*x^{-2} dx$ $\rightarrow$ $\frac{1}{4} \int_\frac{1}{2}^2 -10x-4x^{-2} +21dx$
* Now we can split this into multiple integrals:

$\frac{1}{4}(\int_\frac{1}{2}^2 -10xdx+\int_\frac{1}{2}^2 -4x^{-2}dx+\int_\frac{1}{2}^2 21dx)$
* Taking antiderivatives we get:

$(\frac{1}{4}-10)(\frac{1}{2}x^2)+(-4*-x^{-1})+21x$
* Plugging in our integralsof $\frac{1}{2}$ and $2$ our final answer is: $\frac{27}{16}$.

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3. Find the area of the region in the first quadrant enclosed by the curves $y=sinx$, $y=cosx$ and the y-axis.

* First we set $sinx=cosx$ in order to determine the x value for the integral.
$sinx=cosx$ $\rightarrow$ $sinx$ is only equal to $cosx$ at $\frac{pi}{4}$ so $x=\frac{pi}{4}$

$\int_0^\frac{pi}{4} cosxdx-\int_0^\frac{pi}{4} sinxdx$
* Taking the antiderivatives we get:

$sinx$ from $0$ to $\frac{pi}{4}$ and then subtracting $-cosx$ from $0$ to $\frac{pi}{4}$
* Plugging in the integrals we get:

$\frac{1}{\sqrt{2}} -0--(\frac{1}{\sqrt{2}} -1)$ $\rightarrow$ $\frac{2}{\sqrt{2}} +1$
* Final answer is:$\frac{2}{\sqrt{2}} +1$

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4. Consider the area enclosed by the y-axis, the line $y=1$ and the curve $y=x^2$. Find the value c so that the line $y=c$ splits that area in two equal parts.
* We first need to approach this problem in terms of $y$ instead of $x$.
$x=0$,$y=1$ and $x=\sqrt{y}$
* We need to find a point between $0$ and $1$ where $y=c$ divides the area in half.
* We will use the integrals:
$\int_0^c 2\sqrt{y}dx =\int_c^1 2\sqrt{y} dx$
* We set the integrals equal to each other in order to find the value of c.
* Take the antiderivativs:

$2* \frac{2}{3} y^\frac{3}{2} =2*\frac{2}{3} y^\frac{3}{2}$
* Next solve plugging in the integrals $0$ to $c$.

$(2)\frac{2}{3} (c)^\frac{3}{2} -0=(2)\frac{1}{3} (1)^\frac{3}{2} -(2)\frac{2}{3} (c)^\frac{3}{2}$
* Solving for $c$ we get:
$\frac{8c^\frac{3}{2}}{3}-\frac{2}{3}=0$ $\rightarrow$ $\frac{8c^\frac{3}{2}}{3} =\frac{2}{3}$ $\rightarrow$ $8c^\frac{3}{2} =2$ $\rightarrow$ $c^\frac{3}{2} =\frac{1}{4}$ $\rightarrow$ $(c^\frac{3}{2})^\frac{2}{3} =(\frac{1}{4})^\frac{2}{3}$
* The final answer: $c=\frac{1}{4^\frac{2}{3}}$.

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5. Compute the area of the region enclosed by the curve $y=x^2$, the $x$ axis, and the tangent line of the curve $y=x^2$ at the point $(a,a^2)$.
* The tangent line for $y=x^2$ will be:

$y-a^2 =2a(x-a)$
* setting $a=2$ we get:

$y=4x-4$ as the official tangent line.
* The integral will be $\int_0^{a^2}$ so in terms of x we get:

$\int_0^{a^2} \frac{y-a^2}{2a} dy +\int_0^{a^2} a dy -\int_0^{a^2} \sqrt{y} dy$ $\rightarrow$ $\frac{1}{2a} \int_0^{a^2} y-a^2 dy + ay -\frac{2}{3} y^\frac{3}{2}$

$\frac{1}{2a}*(\frac{1}{2}(a^2)-a^2)-(0-a^2)+(a*a^2))-(a*0)-\frac{2}{3} (a^2)^\frac{3}{2}$
* The area is:

$\frac{2a^2}{2} -\frac{2}{3} (a^2)^\frac{3}{2}$

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