What is the net area between #f(x) = x^3+6/x # and the x-axis over #x in [2, 4 ]#?

Question

Luke Alvarez · Accepted Answer

64.1589

Integrate #f(x)# over the region setting 2 as the lower limit and 4 as the upper limit. #Area = int_2^4f(x)dx# #= int_2^4 x^3+6/xdx = [x^4/4+6lnAbs(x)]_2^4# # = {(4)^4/4+6ln(4)}-{(2)^4/4+6ln(2)}# #=64+12ln(2)-4-6ln(2)# #=60+6ln(2) approx 64.1589#

Emily Ammerman · Answer

undefined To find the net area between the function $ f(x) = \frac{x^3 + 6}{x} $ and the x-axis over the interval $ x $ in [2, 4], we need to compute the definite integral of the absolute value of $ f(x) $ over that interval. This is because the function may be both above and below the x-axis, resulting in both positive and negative areas that need to be accounted for.

$ f(x) = \frac{x^3 + 6}{x} $

So, we need to compute:

$$ \int_{2}^{4} |f(x)| \, dx $$

$$ = \int_{2}^{4} \left|\frac{x^3 + 6}{x}\right| \, dx $$

$$ = \int_{2}^{4} \left|x^2 + \frac{6}{x}\right| \, dx $$

$$ = \int_{2}^{4} \left(x^2 + \frac{6}{x}\right) \, dx $$

$$ = \left[\frac{x^3}{3} + 6\ln|x|\right]_{2}^{4} $$

$$ = \left[\frac{4^3}{3} + 6\ln|4|\right] - \left[\frac{2^3}{3} + 6\ln|2|\right] $$

$$ = \left[\frac{64}{3} + 6\ln(4)\right] - \left[\frac{8}{3} + 6\ln(2)\right] $$

$$ = \left[\frac{64}{3} + 6\ln(2^2)\right] - \left[\frac{8}{3} + 6\ln(2)\right] $$

$$ = \left[\frac{64}{3} + 12\ln(2)\right] - \left[\frac{8}{3} + 6\ln(2)\right] $$

$$ = \frac{56}{3} + 6\ln(2) $$

$$ \approx 56.71 $$