Answer
$1548.8$
Work Step by Step
To solve the problem, first we need to create an equation for it. The fact that it calls for the area between a function and the x-axis tells us we'll use integration. The given x values are then are limits.
So we get...
$\int\limits_{2}^{6}(x-4)^4*dx$
Now we do u-substitution.
$u=(x-4)$ and $du=\frac{dx}{1}$ The 1 is found be doing the derivative of (x-4).
Substituting that we get...
$\int\limits_{2}^{6}u^4*du$
Next, we use the other part of the substitution rule for definite integrals for the limits. 2 goes to $(2)-4=-2$ and 6 goes to $(6)-4=2$ so we have
$\int\limits_{-2}^{2}u^4*du$
Now find the antiderivative...
$\int u^4*du=\frac{1}{5}u^5+c$
substitute u back out
$\frac{1}{5}(x-4)^5+c$
so that we can use the fundamental theorem of calculus
$F(b)-F(c)$
$\frac{1}{5}((2)-4)^5+c-(\frac{1}{5}((-2)-4)^5+c)$
$=\frac{1}{5}*-32+c-(\frac{1}{5}*-7776+c)= \frac{-32}{5}-\frac{-7776}{5}=\frac{7744}{5}=1548.8$
