Which value of $a$ maximizes $\int_{a-1}^{a+1}\frac{1}{1+x^{8}}dx$? I am not being able to understand the graphical method of solving this, any simple explanation will be appreciated.
A non-graphical calculation will be very helpful too.
Thank you so much in advance!
 A: Graphical solution
Click here to see the interactive graph. The red line is the graph of $f(x) = \dfrac1{1+x^8}$. The blue region is the integral. You can drag the slider to see how different values of $a$ gives different areas. Graphically, when $a=0$, the blue region is the largest.
Screenshot:


Non-graphical solution
By Leibniz integral rule:
$$\begin{array}{rcl}
f(a) &=& \displaystyle \int_{a-1}^{a+1}\frac{1}{1+x^{8}} \ \mathrm dx \\
f'(a) &=& \displaystyle \frac{1}{1+(a+1)^{8}} \cdot 1 - \frac{1}{1+(a-1)^{8}} \cdot 1 + \int_{a+1}^{a-1}0 \ \mathrm dx \\
&=& \displaystyle \frac{1}{1+(a+1)^{8}} - \frac{1}{1+(a-1)^{8}} \\\\
\displaystyle \frac{1}{1+(a+1)^{8}} - \frac{1}{1+(a-1)^{8}} &=& 0 \\
\displaystyle \frac{1}{1+(a+1)^{8}} &=& \displaystyle \frac{1}{1+(a-1)^{8}} \\
(a+1)^8 &=& (a-1)^8 \\
a+1 &=& \pm(a-1) \\
a+1 &=& -(a-1) \\
2a &=& 0 \\
a &=& 0 \\
\end{array}$$
Use the first derivative test or the second derivative test to confirm that $a=0$ corresponds to the local maximum.
A: @DHMO works out the details. You don't want to integrate this function.
The plot of the function shows it is symmetrical, and strongly centered at the origin.
$$
f(x) = \frac{1}{1+x^{8}}
$$

A: Let's use our Calculus skills.  Define a function $f$ by
$$f(a)=\int_{a-1}^{a+1}\frac{1}{1+t^8}dt$$
You want to minimize $f$?  Well, we should take a derivative and set it equal to zero...can you see the fundamental theorem of calculus in your future?
$$0=f'(a)=\frac{d}{da}\left[\int_{a-1}^0\frac{1}{1+t^8}dt+\int_0^{a+1}\frac{1}{1+t^8}dt\right]$$
Applying the Fundamental Theorem of Calculus twice, we get
$$0=f'(a)=-\frac{1}{1+(a-1)^8}+\frac{1}{1+(a+1)^8}$$
Now it's an algebra problem:
\begin{align}
\frac{1}{1+(a-1)^8}&=\frac{1}{1+(a+1)^8}\\
1+(a-1)^8&=1+(a+1)^8\\
(a-1)^8&=(a+1)^8\\
\pm(a-1)&=a+1
\end{align}
And we see that $a=0$.  There's your local extreme point.  Probably the easiest way (without graphing) to see that it's a local minimum is to use the first derivative test.
A: We can apply the fundamental theorem of calculus. Let $f(x)=\frac{1}{1+x^8}$. Then if $F(t)=\int_0^t f(x)dx$, we have $F'(t)=f(t)$. We also have $\int_a^b f(x)dx=F(b)-F(a)$. Putting this together, we can write $$\int_{a-1}^{a+1}f(x)dx=F(a+1)-F(a-1).$$ Taking the derivative with respect to $a$, we get $$F'(a+1)-F'(a-1)=f(a+1)-f(a-1)=\frac{1}{1+(a+1)^8}-\frac{1}{1+(a-1)^8}.$$ We find the critical points by setting this equal to zero, so we end up with $1+(a+1)^8=1+(a-1)^8$, or $(a+1)^8-(a-1)^8=0$. We can apply the binomial theorem to rewrite this as $$\sum_{k=0}^8\binom{8}{k}(1+(-1)^k)a^k=16a^7+112a^5+112a^3+16a=0.$$ We simplify slightly to obtain $a(a^6+7a^4+7a^2+1)=0$. We can check that two more roots are $\pm i$. Factoring leaves us with $$a(a^2+1)(a^4+6a^2+1)=a(a^2+1)\left(a^2-\dfrac{-6+\sqrt{32}}{2}\right)\left(a^2-\dfrac{-6-\sqrt{32}}{2}\right).$$ We can check that all of the factors except $a$ have only imaginary roots, so the only real root is $a=0$. Now, we should check that this critical point is a maximum. We compute the second derivative of the original expression: $$\frac{d}{da}\left(\frac{1}{1+(a+1)^8}-\frac{1}{1+(a-1)^8}\right)=-\frac{8(a+1)^7}{(1+(a+1)^8)^2}+\frac{8(a-1)^7}{(1+(a-1)^8)^2}.$$ Plugging in $a=0$, we obtain $-4$, so $0$ is indeed a local maximum (and since it is the only local maximum on an unbounded domain, it is indeed the global maximum).
