Area Calculation for the Region Bounded by Curve $y = x^5$

To find the area of the region bounded by the curve $y = x^5$, the x-axis, and the vertical lines (ordinates) $x = -1$ and $x = 1$, we need to calculate the definite integral of the function's absolute value over the given interval.

Understanding the Function's Behavior

The function is $y = x^5$. We need to consider its sign within the interval $[-1, 1]$:

• For $x$ in $(-1, 0)$, $x^5$ is negative (e.g., $(-0.5)^5 = -0.03125$).
• For $x$ in $(0, 1)$, $x^5$ is positive (e.g., $(0.5)^5 = 0.03125$).
• At $x = 0$, $y = 0$.

Since the area must always be a positive value, we integrate the absolute value of the function, $|x^5|$.

Setting Up the Definite Integral

The area (A) is given by the integral:

$A = \int_{-1}^{1} |x^5| \, dx$

Because the function changes sign at $x = 0$, we split the integral into two parts:

• From $x = -1$ to $x = 0$, $|x^5| = -x^5$.
• From $x = 0$ to $x = 1$, $|x^5| = x^5$.

So, the integral becomes:

$A = \int_{-1}^{0} (-x^5) \, dx + \int_{0}^{1} x^5 \, dx$

Evaluating the Integrals Step-by-Step

First Integral: Area from -1 to 0

Calculate the integral of $-x^5$ from $-1$ to $0$:

$\int_{-1}^{0} (-x^5) \, dx = \left[ -\frac{x^6}{6} \right]_{-1}^{0}$

Now, substitute the limits:

$= \left( -\frac{(0)^6}{6} \right) - \left( -\frac{(-1)^6}{6} \right)$

$= (0) - \left( -\frac{1}{6} \right)$

$= \frac{1}{6}$

Second Integral: Area from 0 to 1

Calculate the integral of $x^5$ from $0$ to $1$:

$\int_{0}^{1} x^5 \, dx = \left[ \frac{x^6}{6} \right]_{0}^{1}$

Now, substitute the limits:

$= \frac{(1)^6}{6} - \frac{(0)^6}{6}$

$= \frac{1}{6} - 0$

$= \frac{1}{6}$

Calculating the Total Area

Add the results from the two integrals:

$A = \frac{1}{6} + \frac{1}{6}$

$A = \frac{2}{6}$

$A = \frac{1}{3}$

The area of the region bounded by the curve $y = x^5$, the x-axis, and the ordinates $x = -1$ and $x = 1$ is $\frac{1}{3}$ square units.