Answer is: option2
\( \dfrac{1}{4} \)Solution:
Set the two equations equal to each other to find where they intersect:
\( \sqrt[3]{x} = x \)
Cube both sides to eliminate the cube root:
\( x = x^3 \)
Rearrange the equation:
\( x^3 - x = 0 \Rightarrow x(x^2 - 1) = 0 \)
Solve for \( x \):
\( x = 0 \quad \text{or} \quad x = \pm 1 \)
Since we are in the first quadrant, we consider \( x = 0 \) and \( x = 1 \). The points of intersection are \( (0, 0) \) and \( (1, 1) \).
In the interval \([0, 1]\), compare the two functions:
-
For \( x = \dfrac{1}{2} \):
\( \sqrt[3]{\dfrac{1}{2}} \approx 0.7937 \quad \text{and} \quad \dfrac{1}{2} = 0.5 \)
Here, \( \sqrt[3]{x} > x \). -
For \( x = 1 \):
\( \sqrt[3]{1} = 1 \quad \text{and} \quad 1 = 1 \)
They are equal at \( x = 1 \).
Therefore, \( y = \sqrt[3]{x} \) is above \( y = x \) in the interval \([0, 1]\).
The area \( A \) between the two curves from \( x = 0 \) to \( x = 1 \) is:
\( A = \int_0^1 (\sqrt[3]{x} - x)\,dx \)
\( A = \left[ \dfrac{3}{4}x^{4/3} - \dfrac{1}{2}x^2 \right]_0^1 \)
\( = \left( \dfrac{3}{4}(1)^{4/3} - \dfrac{1}{2}(1)^2 \right) - \left( \dfrac{3}{4}(0)^{4/3} - \dfrac{1}{2}(0)^2 \right) = \dfrac{3}{4} - \dfrac{1}{2} = \dfrac{1}{4} \)
The area of the region is \( \boxed{\dfrac{1}{4}} \).
Answer: (B) \( \dfrac{1}{4} \)