Find the area between the curve `y=x(x-3)` and the ordinates x = 0 and x = 5.

Answer

To find the area between the curve \(y = x(x - 3)\) and the ordinates \(x = 0\) and \(x = 5\), we will follow these steps:

### Step 1: Identify the curve and the points of intersection
The curve is given by:
\[ y = x(x - 3) = x^2 - 3x \]
To find the points of intersection with the x-axis, we set \(y = 0\):
\[ x(x - 3) = 0 \]
This gives us the points \(x = 0\) and \(x = 3\).

### Step 2: Determine the area between the curve and the x-axis
We need to calculate the area from \(x = 0\) to \(x = 5\). The curve is below the x-axis from \(x = 0\) to \(x = 3\) and above the x-axis from \(x = 3\) to \(x = 5\). Therefore, we will split the area into two parts:
1. Area from \(x = 0\) to \(x = 3\) (below the x-axis)
2. Area from \(x = 3\) to \(x = 5\) (above the x-axis)

### Step 3: Set up the integrals for the areas
The area \(A\) can be expressed as:
\[ A = \int_0^3 -f(x) \, dx + \int_3^5 f(x) \, dx \]
where \(f(x) = x(x - 3)\).

### Step 4: Calculate the first integral
For the first integral:
\[ \int_0^3 -f(x) \, dx = \int_0^3 -(x^2 - 3x) \, dx = \int_0^3 (-x^2 + 3x) \, dx \]
Calculating this integral:
\[ \int (-x^2 + 3x) \, dx = -\frac{x^3}{3} + \frac{3x^2}{2} \]
Evaluating from \(0\) to \(3\):
\[ \left[-\frac{3^3}{3} + \frac{3 \cdot 3^2}{2}\right] - \left[-\frac{0^3}{3} + \frac{3 \cdot 0^2}{2}\right] = \left[-9 + \frac{27}{2}\right] - 0 = -9 + 13.5 = 4.5 \]

### Step 5: Calculate the second integral
For the second integral:
\[ \int_3^5 f(x) \, dx = \int_3^5 (x^2 - 3x) \, dx \]
Calculating this integral:
\[ \int (x^2 - 3x) \, dx = \frac{x^3}{3} - \frac{3x^2}{2} \]
Evaluating from \(3\) to \(5\):
\[ \left[\frac{5^3}{3} - \frac{3 \cdot 5^2}{2}\right] - \left[\frac{3^3}{3} - \frac{3 \cdot 3^2}{2}\right] \]
Calculating each part:
\[ = \left[\frac{125}{3} - \frac{75}{2}\right] - \left[9 - \frac{27}{2}\right] \]
Finding a common denominator (6):
\[ = \left[\frac{250}{6} - \frac{225}{6}\right] - \left[\frac{54}{6} - \frac{81}{6}\right] \]
This simplifies to:
\[ = \frac{25}{6} - \left[-\frac{27}{6}\right] = \frac{25}{6} + \frac{27}{6} = \frac{52}{6} = \frac{26}{3} \]

### Step 6: Combine the areas
Now, we can combine both areas:
\[ A = 4.5 + \frac{26}{3} \]
Converting \(4.5\) to a fraction:
\[ 4.5 = \frac{9}{2} = \frac{27}{6} \]
Thus:
\[ A = \frac{27}{6} + \frac{52}{6} = \frac{79}{6} \]

### Final Answer
The area between the curve \(y = x(x - 3)\) and the ordinates \(x = 0\) and \(x = 5\) is:
\[ \boxed{\frac{79}{6}} \]