# Presentation Slides ## Question 4 --- 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. <iframe src="https://www.desmos.com/calculator/2goklqmgqo?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe> --- ### We first need to approach this problem in terms of $y$ instead of $x$. $x=0$ which is the $y$ axis, $y=1$ will remain the same $y=x^2$ will change to $\rightarrow$ $x=\sqrt{y}$ --- ### We need to find a point between $0$ and $1$ where $y=c$ divides the area in half. * We will use the integrals: $\int_0^c 2\sqrt{y}dx =\int_c^1 2\sqrt{y} dx$ $\rightarrow$ $2\int_0^c\sqrt{y}dx=2\int_c^1\sqrt{y}dx$ * We use $2\sqrt{y}$ for each intergral because $\sqrt{y}$ only accounts for one side of the parabola. * We set the integrals equal to each other in order to find the value of c. --- ### Take the antiderivatives: $2* \frac{2}{3} y^\frac{3}{2} =2*\frac{2}{3} y^\frac{3}{2}$ * Still using the intervals $0$ to $c$ and $c$ to $1$. --- ## Next solve plugging in the integrals $0$ to $c$ and $c$ to $1$. $(2)\frac{2}{3} (c)^\frac{3}{2} -0=(2)\frac{1}{3} (1)^\frac{3}{2} -(2)\frac{2}{3} (c)^\frac{3}{2}$ * 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: $c=\frac{1}{4^\frac{2}{3}}$. --- <iframe src="https://www.desmos.com/calculator/2goklqmgqo?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>