the boundaries of the shaded region are the y - axis, the line y = 1, and the curve y = \\sqrt3{x}.\nfind…

Explanation:

Step1: Rewrite the function

Given $y = \sqrt[3]{x}$, we can rewrite it as $x=y^{3}$.

Step2: Determine the limits of integration

The region is bounded by $y = 0$ and $y = 1$.

Step3: Set up the integral

The area $A$ between two curves with $x$ written as a function of $y$ is given by $A=\int_{a}^{b}(x_{right}-x_{left})dy$. Here, $x_{right}=y^{3}$ and $x_{left} = 0$, and $a = 0$, $b = 1$. So the integral is $\int_{0}^{1}(y^{3}-0)dy=\int_{0}^{1}y^{3}dy$.

Step4: Evaluate the integral

Using the power - rule for integration $\int y^{n}dy=\frac{y^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\left[\frac{y^{4}}{4}\right]_{0}^{1}$.

Step5: Calculate the definite - integral value

$\frac{1^{4}}{4}-\frac{0^{4}}{4}=\frac{1}{4}$.

Answer:

$\frac{1}{4}$