Question

let the region r be the area enclosed the function $f(x) = x^{\frac{1}{3}} + 2$, the horizontal line $y = 4$, and the $y$-axis. write an integral in terms of $x$ and also an integral in terms of $y$ that would represent the area of the region r. if necessary, round limit values to the nearest thousandth.

answer attempt 1 out of 3
$x_1 = \square$ $x_2 = \square$ $\int_{x_1}^{x_2} \left \square \
ight dx$
$y_1 = \square$ $y_2 = \square$ $\int_{y_1}^{y_2} \left \square \
ight dy$

Explanation:

Step1: Find intersection point (x - terms)

Set \(f(x)=x^{\frac{1}{3}}+2 = 4\). Then \(x^{\frac{1}{3}}=2\), so \(x = 8\).
For the integral in terms of \(x\), the lower limit \(x_1 = 0\) (from \(y -\)axis, \(x = 0\)) and upper limit \(x_2=8\). The top function is \(y = 4\) and the bottom function is \(y=x^{\frac{1}{3}}+2\). The integrand is \(4-(x^{\frac{1}{3}}+2)=2 - x^{\frac{1}{3}}\).

Step2: Rewrite function for \(y -\)terms

Solve \(y=x^{\frac{1}{3}}+2\) for \(x\): \(x=(y - 2)^3\).
For the integral in terms of \(y\), the lower limit \(y_1 = 2\) (when \(x = 0\), \(y=x^{\frac{1}{3}}+2=2\)) and upper limit \(y_2 = 4\). The right - hand function with respect to \(y\) is \(x=(y - 2)^3\) (from \(y=x^{\frac{1}{3}}+2\)) and the left - hand function is \(x = 0\). The integrand is \((y - 2)^3-0=(y - 2)^3\).

Answer:

For the integral in terms of \(x\): \(x_1 = 0\), \(x_2=8\), \(\int_{0}^{8}(2 - x^{\frac{1}{3}})dx\)
For the integral in terms of \(y\): \(y_1 = 2\), \(y_2 = 4\), \(\int_{2}^{4}(y - 2)^3dy\)