Question

The area (in sq. units) of the region bounded by the curve \( y = x^5 \), the x-axis and the ordinates x = -1 and x = 1 is equal to

• A. $\frac{1}{6}$
• B. 1
• C. $\frac{1}{2}$
  (D) $\frac{2}{3}$
• D.

Hint:

For functions that are symmetric about the origin (odd functions like $x^3$, $x^5$, sin(x)), the area from -a to a is twice the area from 0 to a. So, $A = 2 \int_{0}^{a} f(x) dx$. This simplifies the calculation. For this problem, $A = 2 \int_{0}^{1} x^3 dx = 2[\frac{x^4}{4}]_{0}^{1} = 2(\frac{1}{4}) = \frac{1}{2}$.

Answer 1

The Correct Option is C

Note: The question as stated, with \( y = x^5 \), results in an answer of \( \frac{1}{3} \), which is not an available option. It is probable that the intended function was \( y = x^3 \), given the visual similarity between handwritten '3' and '5' and that \( \frac{1}{2} \) (the correct answer for \( y = x^3 \)) is among the choices. This rephrasing proceeds under the assumption that \( y = x^3 \) was the intended function.

Step 1: Concept Clarification:

The objective is to determine the area enclosed by the curve \( y = x^3 \) and the x-axis. As area is inherently a non-negative value, the integral must consider the absolute value of the function, \( |f(x)| \). The function \( y = x^3 \) is negative for \( x<0 \) and positive for \( x>0 \).

Step 2: Fundamental Formula/Methodology:

The area \( A \) is calculated via the definite integral of \( |x^3| \) from -1 to 1: \[ A = \int_{-1}^{1} |x^3| \, dx \] The integral needs to be segmented at \( x = 0 \), the point where the function's sign changes.

Step 3: Detailed Calculation:

The integral is divided into two segments:

\[ A = \int_{-1}^{0} |x^3| \, dx + \int_{0}^{1} |x^3| \, dx \] For \( x \) values in the interval \( [-1, 0) \), \( x^3 \) is negative, thus \( |x^3| = -x^3 \).
For \( x \) values in the interval \( [0, 1] \), \( x^3 \) is positive, thus \( |x^3| = x^3 \).
This leads to the integral setup: \[ A = \int_{-1}^{0} (-x^3) \, dx + \int_{0}^{1} (x^3) \, dx \] Integration yields: \[ A = \left[ -\frac{x^4}{4} \right]_{-1}^{0} + \left[ \frac{x^4}{4} \right]_{0}^{1} \] Evaluating the first segment: \[ \left( -\frac{0^4}{4} \right) - \left( -\frac{(-1)^4}{4} \right) = 0 - \left( -\frac{1}{4} \right) = \frac{1}{4} \] Evaluating the second segment: \[ \left( \frac{1^4}{4} \right) - \left( \frac{0^4}{4} \right) = \frac{1}{4} - 0 = \frac{1}{4} \] Summing the results of both segments to find the total area: \[ A = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \]

Step 4: Conclusive Result:

Based on the assumption that the intended curve was \( y = x^3 \), the calculated area is \( \frac{1}{2} \) square units.