Find integral from 0 to 4 x dx using the limit of Riemann

Find $\int_0^4 x \, dx$ using the limit of Riemann sums (divide $[0,4]$ into $n$ equal parts and take $\xi_i = x_i$), and interpret the number geometrically.

We will evaluate the integral using the definition as the limit of Riemann sums.

Step 1: Partition the interval

Divide the interval $[0,4]$ into $n$ equal subintervals. The width of each subinterval is:

\[\Delta x = \frac{4-0}{n} = \frac{4}{n}\]

The partition points are:

\[x_i = 0 + i \cdot \Delta x = \frac{4i}{n}, \quad i = 0, 1, 2, \ldots, n\]

Step 2: Choose the sample points

We are given that $\xi_i = x_i$, so we use the right endpoint of each subinterval:

\[\xi_i = x_i = \frac{4i}{n}\]

Step 3: Form the Riemann sum

The Riemann sum is:

\[S_n = \sum_{i=1}^{n} f(\xi_i) \Delta x = \sum_{i=1}^{n} \xi_i \cdot \Delta x = \sum_{i=1}^{n} \frac{4i}{n} \cdot \frac{4}{n}\]

\[= \sum_{i=1}^{n} \frac{16i}{n^2} = \frac{16}{n^2} \sum_{i=1}^{n} i\]

Step 4: Use the formula for the sum of integers

Recall that:

\[\sum_{i=1}^{n} i = \frac{n(n+1)}{2}\]

Therefore:

\[S_n = \frac{16}{n^2} \cdot \frac{n(n+1)}{2} = \frac{16n(n+1)}{2n^2} = \frac{8(n+1)}{n}\]

\[= \frac{8n + 8}{n} = 8 + \frac{8}{n}\]

Step 5: Take the limit as $n \to \infty$

\[\int_0^4 x \, dx = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(8 + \frac{8}{n}\right) = 8 + 0 = 8\]

Verification using antiderivative:

\[\int_0^4 x \, dx = \left[\frac{x^2}{2}\right]_0^4 = \frac{16}{2} - 0 = 8\]

Geometric interpretation:

The number $\int_0^4 x \, dx = 8$ represents the area under the curve $y = x$ from $x = 0$ to $x = 4$. This is the area of a right triangle with base 4 and height 4, which equals:

\[\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 4 \cdot 4 = 8\]

Final answer:

\[\boxed{\int_0^4 x \, dx = 8}\]

The integral represents the area of the triangular region bounded by the line $y = x$, the $x$-axis, and the vertical line $x = 4$.