What is the area between the curve f(x) = x |x| and x-axis for x = [-1, 1]?
Understanding the Area Between a Curve and the X-axis
The question asks for the area between the curve defined by the function \(f(x) = x |x|\) and the x-axis over the specific interval \(x \in [-1, 1]\). Finding the area between a curve and the x-axis involves using definite integrals. When the function crosses the x-axis within the given interval, or when the function is defined differently over sub-intervals due to absolute values, we need to be careful. The area is always a positive value, so we calculate the integral for parts where the function is above or below the x-axis separately and sum their absolute values.
Defining the Function \(f(x) = x |x|\)
The function \(f(x) = x |x|\) is defined using the absolute value function. The absolute value \(|x|\) changes its definition depending on whether \(x\) is positive or negative.
- If \(x \ge 0\), then \(|x| = x\). So, for \(x \ge 0\), \(f(x) = x \cdot x = x^2\).
- If \(x < 0\), then \(|x| = -x\). So, for \(x < 0\), \(f(x) = x \cdot (-x) = -x^2\).
Thus, the function \(f(x)\) can be written as a piecewise function:
\(f(x) = \begin{cases} x^2 & \text{if } x \ge 0 \\ -x^2 & \text{if } x < 0 \end{cases}\)
Calculating the Area over the Interval [-1, 1]
The given interval is \([-1, 1]\). This interval includes both negative values (from -1 up to 0) and positive values (from 0 up to 1). Since the definition of \(f(x)\) changes at \(x=0\), we need to split the integral into two parts:
- From \(x = -1\) to \(x = 0\), where \(f(x) = -x^2\).
- From \(x = 0\) to \(x = 1\), where \(f(x) = x^2\).
The total area is the sum of the absolute values of the integrals over these sub-intervals.
Total Area = \(|\int_{-1}^0 f(x) dx| + |\int_{0}^1 f(x) dx|\)
Let's calculate each integral:
Area Calculation for the Interval [-1, 0]
In the interval \([-1, 0]\), \(f(x) = -x^2\). Let's calculate the definite integral:
\(\int_{-1}^0 -x^2 dx\)
The integral of \(-x^2\) is \(-\frac{x^3}{3}\).
\(\int_{-1}^0 -x^2 dx = [-\frac{x^3}{3}]_{-1}^0\)
Now, we evaluate the antiderivative at the limits of integration:
\(= (-\frac{(0)^3}{3}) - (-\frac{(-1)^3}{3})\)
\(= 0 - (-\frac{-1}{3})\)
\(= 0 - \frac{1}{3}\)
\(= -\frac{1}{3}\)
The integral is \(-1/3\). Since the function \(f(x) = -x^2\) is negative over \([-1, 0]\) (except at \(x=0\)), the integral is negative. The area is the absolute value of this result.
Area 1 = \(|-\frac{1}{3}| = \frac{1}{3}\)
Area Calculation for the Interval [0, 1]
In the interval \([0, 1]\), \(f(x) = x^2\). Let's calculate the definite integral:
\(\int_{0}^1 x^2 dx\)
The integral of \(x^2\) is \(\frac{x^3}{3}\).
\(\int_{0}^1 x^2 dx = [\frac{x^3}{3}]_{0}^1\)
Now, we evaluate the antiderivative at the limits of integration:
\(= (\frac{(1)^3}{3}) - (\frac{(0)^3}{3})\)
\(= \frac{1}{3} - 0\)
\(= \frac{1}{3}\)
The integral is \(1/3\). Since the function \(f(x) = x^2\) is positive over \([0, 1]\) (except at \(x=0\)), the integral is positive, and this value directly represents the area.
Area 2 = \(\frac{1}{3}\)
Total Area Between the Curve and X-axis
The total area is the sum of Area 1 and Area 2.
Total Area = Area 1 + Area 2
Total Area = \(\frac{1}{3} + \frac{1}{3}\)
Total Area = \(\frac{2}{3}\)
So, the total area between the curve \(f(x) = x |x|\) and the x-axis for \(x = [-1, 1]\) is \(\frac{2}{3}\).
| Interval | Function \(f(x)\) | Integral \(\int f(x) dx\) | Definite Integral Result | Area |
|---|---|---|---|---|
| \([-1, 0]\) | \(-x^2\) | \(-\frac{x^3}{3}\) | \(-\frac{1}{3}\) | \(|-\frac{1}{3}| = \frac{1}{3}\) |
| \([0, 1]\) | \(x^2\) | \(\frac{x^3}{3}\) | \(\frac{1}{3}\) | \(|\frac{1}{3}| = \frac{1}{3}\) |
Summary of Area Calculation
We split the integration interval \([-1, 1]\) into \([-1, 0]\) and \([0, 1]\) because the function \(f(x) = x|x|\) changes its definition at \(x=0\). For each sub-interval, we calculated the definite integral and took its absolute value to find the area above or below the x-axis. Summing these areas gives the total area.
Revision Table: Key Concepts
| Concept | Description | Relevance to Problem |
|---|---|---|
| Area under Curve | Calculated by integrating \(f(x)\) over an interval \([a, b]\). If \(f(x)\) is sometimes negative, the integral gives net signed area. For total area, use \(|\int f(x) dx|\) for parts where \(f(x)\) is below x-axis, or integrate \(|f(x)|\). | Required to find the area between \(f(x) = x|x|\) and the x-axis. |
| Absolute Value Function \(|x|\) | Defined as \(x\) for \(x \ge 0\) and \(-x\) for \(x < 0\). | Crucial for understanding and splitting the function \(f(x) = x|x|\) into piecewise parts. |
| Definite Integral | Represents the net signed area between a function and the x-axis over a closed interval. Calculated using the Fundamental Theorem of Calculus: \(\int_a^b f(x) dx = F(b) - F(a)\), where \(F(x)\) is an antiderivative of \(f(x)\). | The primary tool used to calculate the area over each sub-interval. |
| Piecewise Function | A function defined by multiple sub-functions, each applying to a different interval of the domain. | \(f(x) = x|x|\) is a piecewise function, requiring the integration interval to be split. |
Additional Information: Understanding \(f(x) = x|x|\) Graph
The graph of \(f(x) = x|x|\) consists of two parts:
- For \(x \ge 0\), the graph is \(y = x^2\), which is the right half of a parabola opening upwards, starting at (0,0).
- For \(x < 0\), the graph is \(y = -x^2\), which is the left half of a parabola opening downwards, also starting at (0,0).
The graph looks like a cubic function (\(y=x^3\)) visually, but it's composed of parabolic segments. For the interval \([-1, 1]\), the graph goes from \(f(-1) = -(-1)^2 = -1\) up to \(f(0)=0\) and then up to \(f(1) = 1^2 = 1\).
- From \(x=-1\) to \(x=0\), the curve is below the x-axis (parabola \(y=-x^2\)). The area here is \(\int_{-1}^0 |-x^2| dx = \int_{-1}^0 x^2 dx\).
- From \(x=0\) to \(x=1\), the curve is above the x-axis (parabola \(y=x^2\)). The area here is \(\int_{0}^1 |x^2| dx = \int_{0}^1 x^2 dx\).
Calculating \(\int x^2 dx\) is straightforward, leading to \(\frac{x^3}{3}\). Evaluating \(\int_{-1}^0 x^2 dx = [\frac{x^3}{3}]_{-1}^0 = 0 - (-\frac{1}{3}) = \frac{1}{3}\). Evaluating \(\int_{0}^1 x^2 dx = [\frac{x^3}{3}]_{0}^1 = \frac{1}{3} - 0 = \frac{1}{3}\). The total area is \(\frac{1}{3} + \frac{1}{3} = \frac{2}{3}\). This confirms the result obtained by integrating \(f(x)\) and taking absolute values.
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