Answer :
To find the area bounded by the curve \(y = x|x|\), the \(x\)-axis, and the ordinates \(x = 1\) and \(x = -1\), we can break the analysis into two segments: one for \(x\) in the interval \([-1, 0]\) and one for \(x\) in the interval \([0, 1]\).
### For \(0 \le x \le 1\):
In this interval, \(x\) is non-negative, so \(|x| = x\). Therefore, the equation \(y = x|x|\) simplifies to:
[tex]\[ y = x \cdot x = x^2. \][/tex]
We need to find the area under the curve \(y = x^2\) from \(x = 0\) to \(x = 1\). We do this by integrating \(x^2\) with respect to \(x\):
[tex]\[ \text{Area}_{\text{positive}} = \int_{0}^{1} x^2 \, dx. \][/tex]
Evaluating this integral:
[tex]\[ \int_{0}^{1} x^2 \, dx = \left[\frac{x^3}{3}\right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}. \][/tex]
### For \(-1 \le x < 0\):
In this interval, \(x\) is negative, so \(|x| = -x\). Therefore, the equation \(y = x|x|\) simplifies to:
[tex]\[ y = x \cdot (-x) = -x^2. \][/tex]
We need to find the area under the curve \(y = -x^2\) from \(x = -1\) to \(x = 0\). We do this by integrating \(-x^2\) with respect to \(x\):
[tex]\[ \text{Area}_{\text{negative}} = \int_{-1}^{0} -x^2 \, dx. \][/tex]
Evaluating this integral:
[tex]\[ \int_{-1}^{0} -x^2 \, dx = -\int_{-1}^{0} x^2 \, dx = -\left[\frac{x^3}{3}\right]_{-1}^{0} = -\left(\frac{0^3}{3} - \frac{(-1)^3}{3}\right) = -\left(0 - \left(-\frac{1}{3}\right)\right) = -\left(0 + \frac{1}{3}\right) = -\frac{1}{3}. \][/tex]
### Total Area Bounded:
The total area bounded by the curve and the \(x\)-axis in the interval \([-1, 1]\) is obtained by adding the absolute values of the areas from the two segments:
[tex]\[ \text{Total Area} = \text{Area}_{\text{positive}} + \text{Area}_{\text{negative}} = \frac{1}{3} + \left(-\frac{1}{3}\right) = 0. \][/tex]
Hence, the area bounded by the curve \(y = x|x|\), the \(x\)-axis, and the ordinates \(x = 1\) and \(x = -1\) is:
[tex]\( \boxed{0} \)[/tex].
### For \(0 \le x \le 1\):
In this interval, \(x\) is non-negative, so \(|x| = x\). Therefore, the equation \(y = x|x|\) simplifies to:
[tex]\[ y = x \cdot x = x^2. \][/tex]
We need to find the area under the curve \(y = x^2\) from \(x = 0\) to \(x = 1\). We do this by integrating \(x^2\) with respect to \(x\):
[tex]\[ \text{Area}_{\text{positive}} = \int_{0}^{1} x^2 \, dx. \][/tex]
Evaluating this integral:
[tex]\[ \int_{0}^{1} x^2 \, dx = \left[\frac{x^3}{3}\right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}. \][/tex]
### For \(-1 \le x < 0\):
In this interval, \(x\) is negative, so \(|x| = -x\). Therefore, the equation \(y = x|x|\) simplifies to:
[tex]\[ y = x \cdot (-x) = -x^2. \][/tex]
We need to find the area under the curve \(y = -x^2\) from \(x = -1\) to \(x = 0\). We do this by integrating \(-x^2\) with respect to \(x\):
[tex]\[ \text{Area}_{\text{negative}} = \int_{-1}^{0} -x^2 \, dx. \][/tex]
Evaluating this integral:
[tex]\[ \int_{-1}^{0} -x^2 \, dx = -\int_{-1}^{0} x^2 \, dx = -\left[\frac{x^3}{3}\right]_{-1}^{0} = -\left(\frac{0^3}{3} - \frac{(-1)^3}{3}\right) = -\left(0 - \left(-\frac{1}{3}\right)\right) = -\left(0 + \frac{1}{3}\right) = -\frac{1}{3}. \][/tex]
### Total Area Bounded:
The total area bounded by the curve and the \(x\)-axis in the interval \([-1, 1]\) is obtained by adding the absolute values of the areas from the two segments:
[tex]\[ \text{Total Area} = \text{Area}_{\text{positive}} + \text{Area}_{\text{negative}} = \frac{1}{3} + \left(-\frac{1}{3}\right) = 0. \][/tex]
Hence, the area bounded by the curve \(y = x|x|\), the \(x\)-axis, and the ordinates \(x = 1\) and \(x = -1\) is:
[tex]\( \boxed{0} \)[/tex].
