Evaluate the following integrals using limit of sum.
`int_(a)^(b)x^(3)dx`

Explanation

To evaluate the integral \( \int_{a}^{b} x^3 \, dx \) using the limit of sums, we can follow these steps:

### Step 1: Define the interval and partition it
We start by partitioning the interval \([a, b]\) into \(n\) equal subintervals. The width of each subinterval is given by:
\[ h = \frac{b - a}{n} \]
The points in the partition can be defined as:
\[ x_r = a + rh \quad \text{for } r = 0, 1, 2, \ldots, n \]

### Step 2: Write the Riemann sum
The Riemann sum for the function \(f(x) = x^3\) over the interval \([a, b]\) is:
\[ S_n = \sum_{r=0}^{n-1} f(x_r) \cdot h = \sum_{r=0}^{n-1} (a + rh)^3 \cdot h \]
Substituting \(f(x)\):
\[ S_n = \sum_{r=0}^{n-1} (a + rh)^3 \cdot h \]

### Step 3: Expand the function
We expand \((a + rh)^3\):
\[ (a + rh)^3 = a^3 + 3a^2(rh) + 3a(rh)^2 + (rh)^3 \]
Thus, we can rewrite the sum:
\[ S_n = \sum_{r=0}^{n-1} \left( a^3 + 3a^2(rh) + 3a(rh)^2 + (rh)^3 \right) \cdot h \]

### Step 4: Distribute \(h\) and separate the sums
Distributing \(h\) gives:
\[ S_n = \sum_{r=0}^{n-1} \left( a^3h + 3a^2rh^2 + 3ar^2h^3 + r^3h^4 \right) \]
This can be separated into individual sums:
\[ S_n = a^3h \sum_{r=0}^{n-1} 1 + 3a^2h^2 \sum_{r=0}^{n-1} r + 3ah^3 \sum_{r=0}^{n-1} r^2 + h^4 \sum_{r=0}^{n-1} r^3 \]

### Step 5: Evaluate the sums
Using the formulas for the sums:
- \( \sum_{r=0}^{n-1} 1 = n \)
- \( \sum_{r=0}^{n-1} r = \frac{(n-1)n}{2} \)
- \( \sum_{r=0}^{n-1} r^2 = \frac{(n-1)n(2n-1)}{6} \)
- \( \sum_{r=0}^{n-1} r^3 = \left( \frac{(n-1)n}{2} \right)^2 \)

Substituting these into \(S_n\):
\[ S_n = a^3h n + 3a^2h^2 \cdot \frac{(n-1)n}{2} + 3ah^3 \cdot \frac{(n-1)n(2n-1)}{6} + h^4 \cdot \left( \frac{(n-1)n}{2} \right)^2 \]

### Step 6: Factor out \(h\) and take the limit as \(n \to \infty\)
Now, we can factor out \(h\) from each term and substitute \(h = \frac{b-a}{n}\):
\[ S_n = \left( \frac{b-a}{n} \right) \left( a^3 n + \frac{3a^2(b-a)(n-1)}{2} + \frac{3a(b-a)^2(n-1)(2n-1)}{6} + \frac{(b-a)^3(n-1)^2}{4} \right) \]
Taking the limit as \(n \to \infty\):
\[ \lim_{n \to \infty} S_n = \int_{a}^{b} x^3 \, dx = \frac{(b^4 - a^4)}{4} \]

### Final Result
Thus, the value of the integral is:
\[ \int_{a}^{b} x^3 \, dx = \frac{b^4 - a^4}{4} \]