AP.6.1.1 A region is bounded between the graphs

[karush]

$\tiny{AP.6.1.1}\\$
$\textsf{Let $f(x)=x^3$}\\$
$\textsf{A region is bounded between the graphs of $y=-1$ and $y=\ f(x)$ }\\$
$\textsf{for x between $-1$ and $0$, region. }\\$
$\textsf{And between the graph of $y=1$ and $y=f(x)$ for x between $0$ and $1$ }\\$
$\textsf{This appears to be symmetrical so in order to get complete area one is found then doubled} \\$
\begin{align}
\displaystyle
I&=2 \int^1_0{(1-x^3)}d{x}\\
&=2 \left[ x-\frac{x^4}{4} \right]_0^1\\
&=2\left[\left[1-\frac{1}{4}\right]_0-\left[0-\frac{0}{4}\right]^1\right] \\
&=2\left[\frac{3}{4}-0\right]\\
&=\frac{6}{4}=\frac{3}{2}
\end{align}
$\textit{think this is ok but suggestions?}$

 

[Greg]

Looks good to me and that is certainly an efficient way of going about it, if not the most efficient. You may want to include a detailed note as to why you are multiplying the integral by 2, using the fact that $x^3$ is an odd function. :)

 

[karush]

the areas for given interval's are the same but one is positive and the other is negative so they cancel, so just multiply the positive one twice. may not be conventional though..
r$\tiny{s6.12.13}$
$\textsf{Find an equation of the sphere}\\$
$\textsf{that passes through the point (4,3,-1) and has center (3,8,1)} $
\begin{align}
\displaystyle
(x-3)^2+(y-8)^2+(z-1)^2&= r^2\\
\sqrt{(3-4)^2+(8-3)^2+(1+1)^2}&=r^2\\
\sqrt{1+25+4}&=\sqrt{30}^2=30 =r
\end{align}
$\textit{so far ??}$

 

[MarkFL]

the areas for given interval's are the same but one is positive and the other is negative so they cancel, so just multiply the positive one twice. may not be conventional though..

If you were computing the integral, then yes the odd-function rule would immediately tell you it evaluates to zero. However, you are asked to compute a bounded area, so that's a bit different. You could state:

\(\displaystyle A=\int_{-1}^0 x^3-(-1)\,dx+\int_{0}^1 1-x^3\,dx=\int_{-1}^0 1+x^3\,dx+\int_{0}^1 1-x^3\,dx\)

Now, for the first integral on the RHS, let's let:

\(\displaystyle u=-x\implies du=-dx\)

And for the second integral, let's let:

\(\displaystyle u=x\implies du=dx\)

Now, for the first integral we can replace the negative sign on the differential with a change of limits (and the limits get negated as per the substitution), and after we make the substitutions, we have:

\(\displaystyle A=\int_{0}^{1} 1-u^3\,du+\int_{0}^1 1-u^3\,du=2\int_{0}^1 1-u^3\,du\)