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 provided question with \( y = x^5 \) yields an answer of \( \frac{1}{3} \), which is not among the options. It is highly likely that the question intended to ask for \( y = x^3 \), as handwritten '3' and '5' can look similar, and the correct answer for \( y = x^3 \) is one of the options. We will proceed with this assumption.

Step 1: Understanding the Concept:

We need to find the area bounded by a curve and the x-axis. Since area is always a non-negative quantity, we must integrate 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: Key Formula or Approach:

The area \( A \) is given by the definite integral of \( |x^3| \) from -1 to 1: \[ A = \int_{-1}^{1} |x^3| \, dx \] We must split the integral at \( x = 0 \) where the function's sign changes.

Step 3: Detailed Explanation:

Let's split the integral into two parts:

\[ A = \int_{-1}^{0} |x^3| \, dx + \int_{0}^{1} |x^3| \, dx \] For \( x \in [-1, 0) \), \( x^3 \) is negative, so \( |x^3| = -x^3 \).
For \( x \in [0, 1] \), \( x^3 \) is positive, so \( |x^3| = x^3 \).
\[ A = \int_{-1}^{0} (-x^3) \, dx + \int_{0}^{1} (x^3) \, dx \] Now, we integrate: \[ A = \left[ -\frac{x^4}{4} \right]_{-1}^{0} + \left[ \frac{x^4}{4} \right]_{0}^{1} \] Evaluate the first part: \[ \left( -\frac{0^4}{4} \right) - \left( -\frac{(-1)^4}{4} \right) = 0 - \left( -\frac{1}{4} \right) = \frac{1}{4} \] Evaluate the second part: \[ \left( \frac{1^4}{4} \right) - \left( \frac{0^4}{4} \right) = \frac{1}{4} - 0 = \frac{1}{4} \] The total area is the sum of the two parts: \[ A = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \]

Step 4: Final Answer:

Assuming the intended curve was \( y = x^3 \), the area is \( \frac{1}{2} \) square units.