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.
Important Questions from Application of Integrals
What is the area of the region in the first quadrant enclosed by the x-axis the line \({\rm{x}} = \sqrt 3 \) and the circle?
What is the area of the region in the first quadrant enclosed by the x-axis the line \({\rm{x}} = \sqrt 3 {\rm{\;y\;}}\) and the circle?
What is the area of the region bounded by the above two curves and the lines x = 0 and \({\rm{x}} = \frac{{\rm{\pi }}}{4}\) ?
What is the area of the region bounded by the above two curves and the lines \({\rm{x}} = \frac{{\rm{\pi }}}{4}\) and \(= \frac{{\rm{\pi }}}{2}\) ?
The area bounded by the coordinate axes and the curve \(\sqrt {\rm{x}} + \sqrt {\rm{y}} = 1\) , is
