Question

The area of the region bounded by \( y = x^{5/2} \) and \( y = x \) (in square units) is

• A. \( \frac{3}{7} \)
• B. \( \frac{2}{7} \)
• C. \( \frac{3}{14} \)
• D. \( \frac{5}{14} \)
• E. \( \frac{4}{7} \)

Hint:

Always check which curve is above before integrating.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
To find the area of a region bounded by two curves, we first need to find their points of intersection. Then, we integrate the difference between the upper curve and the lower curve over the interval defined by these intersection points.
Step 2: Key Formula or Approach:
1. Find the points of intersection by setting the two equations equal to each other: \(x^{5/2} = x\). 2. Determine which function is the "upper" curve and which is the "lower" curve in the interval between the intersection points. 3. The area `A` is given by the definite integral: \(A = \int_{a}^{b} (y_{\text{upper}} - y_{\text{lower}}) dx\), where `a` and `b` are the x-coordinates of the intersection points.
Step 3: Detailed Explanation:
1. Find the intersection points. \[ x^{5/2} = x \] \[ x^{5/2} - x = 0 \] \[ x(x^{3/2} - 1) = 0 \] This gives two solutions: \(x = 0\) and \(x^{3/2} = 1 \implies x = 1\). So, the curves intersect at \(x=0\) and \(x=1\). These will be our limits of integration. 2. Determine the upper and lower curves. Let's pick a test point in the interval (0, 1), for example, \(x = 1/4\).
For \(y=x\): \(y = 1/4 = 0.25\)
For \(y=x^{5/2}\): \(y = (1/4)^{5/2} = (\sqrt{1/4})^5 = (1/2)^5 = 1/32 \approx 0.03125\)
Since \(1/4>1/32\), the line \(y=x\) is the upper curve and \(y=x^{5/2}\) is the lower curve in the interval [0, 1]. 3. Set up and evaluate the integral. \[ A = \int_{0}^{1} (x - x^{5/2}) dx \] \[ A = \left[ \frac{x^2}{2} - \frac{x^{5/2 + 1}}{5/2 + 1} \right]_{0}^{1} \] \[ A = \left[ \frac{x^2}{2} - \frac{x^{7/2}}{7/2} \right]_{0}^{1} = \left[ \frac{x^2}{2} - \frac{2}{7}x^{7/2} \right]_{0}^{1} \] Now, apply the limits of integration: \[ A = \left( \frac{1^2}{2} - \frac{2}{7}(1)^{7/2} \right) - \left( \frac{0^2}{2} - \frac{2}{7}(0)^{7/2} \right) \] \[ A = \left( \frac{1}{2} - \frac{2}{7} \right) - 0 \] \[ A = \frac{7 - 4}{14} = \frac{3}{14} \] Step 4: Final Answer:
The area of the region is \(\frac{3}{14}\) square units.