Maximum value of the integral

Question

The maximum value of the integral

$$\int_{a-1}^{a+1} \frac{1}{1+x^8}dx $$

is attained

A) at exactly two values of $a$

B) only at one value of $a$, which is positive

C) only at one value of $a$ which is negative

D) only at $a=0$

Answer 1

Differentiate it and set it equal to zero to get

$$(a-1)^8 = (a+1)^8 \implies 1-a=a+1 \implies a=0$$

Answer 2

By inspection $f(x)=\frac{1}{1+x^8}$ has a maximum value at $x=0$, is symmetric about the $y$ axis and decreases monotonically away from that axis. Any interval of fixed width on the $x$ axis will include the maximum area under the graph $f(x)$ if the interval is centred on $x=0$. The only interval from $x=a-1$ to $x=a+1$ which is centred on $x=0$ is that with $a=0$.