Which value of $a$ maximizes $\int_{a-1}^{a+1}\frac{1}{1+x^{8}}dx$?

Question

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!

Answer 1

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:

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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.

Answer 2

@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.

$$ f(x) = \frac{1}{1+x^{8}} $$

Answer 3

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.

Answer 4

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).