## Calculus: Early Transcendentals (2nd Edition)

$1548.8$
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 $F(b)-F(c)$ $\frac{1}{5}((2)-4)^5+c-(\frac{1}{5}((-2)-4)^5+c)$ $=\frac{1}{5}*-32+c-(\frac{1}{5}*-7776+c)= \frac{-32}{5}-\frac{-7776}{5}=\frac{7744}{5}=1548.8$