To express the given integrals as a limit of a Riemann sum, we'll break down the general approach for each one.

1. $\int_0^1 (1 + x^2) dx$

A Riemann sum for an integral $\int_a^b f(x) dx$ is given by the limit:

$\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x$

Where:

• $\Delta x = \frac{b - a}{n}$, which represents the width of each subinterval.
• $x_i^* = a + i \Delta x$, which is a sample point in the $i$-th subinterval (it could be left, right, or mid-point, but typically the mid-point is used unless otherwise specified).

For the integral $\int_0^1 (1 + x^2) dx$:

• $a = 0$, $b = 1$, and $f(x) = 1 + x^2$.
• $\Delta x = \frac{1 - 0}{n} = \frac{1}{n}$.
• $x_i^* = 0 + i \Delta x = \frac{i}{n}$.

So the Riemann sum approximation becomes:

$S_n = \sum_{i=1}^{n} \left( 1 + \left(\frac{i}{n}\right)^2 \right) \frac{1}{n}$

The integral is the limit as $n \to \infty$:

$\int_0^1 (1 + x^2) dx = \lim_{n \to \infty} \sum_{i=1}^{n} \left( 1 + \left(\frac{i}{n}\right)^2 \right) \frac{1}{n}$

Optional: Evaluating the Integral

To evaluate $\int_0^1 (1 + x^2) dx$, we can compute the integral directly:

$\int_0^1 (1 + x^2) dx = \left[ x + \frac{x^3}{3} \right]_0^1 = (1 + \frac{1^3}{3}) - (0 + \frac{0^3}{3}) = 1 + \frac{1}{3} = \frac{4}{3}$

2. $\int_1^3 (1 - x) dx$

For the second integral, $\int_1^3 (1 - x) dx$, we follow the same procedure:

• $a = 1$, $b = 3$, and $f(x) = 1 - x$.
• $\Delta x = \frac{3 - 1}{n} = \frac{2}{n}$.
• $x_i^* = 1 + i \Delta x = 1 + \frac{2i}{n}$.

The Riemann sum becomes:

$S_n = \sum_{i=1}^{n} \left( 1 - \left(1 + \frac{2i}{n}\right) \right) \frac{2}{n} = \sum_{i=1}^{n} \left( - \frac{2i}{n} \right) \frac{2}{n}$

The integral is the limit as $n \to \infty$:

$\int_1^3 (1 - x) dx = \lim_{n \to \infty} \sum_{i=1}^{n} \left(1 - \left(1 + \frac{2i}{n}\right) \right) \frac{2}{n}$

Optional: Evaluating the Integral

To evaluate $\int_1^3 (1 - x) dx$, we compute the integral:

$\int_1^3 (1 - x) dx = \left[ x - \frac{x^2}{2} \right]_1^3 = \left(3 - \frac{9}{2}\right) - \left(1 - \frac{1}{2}\right) = 3 - 4.5 - (1 - 0.5) = -2$

Would you like more details or have any questions?

Here are five related questions to deepen your understanding:

1. How would the Riemann sum change if we used midpoints instead of right endpoints?
2. What is the geometric interpretation of the definite integrals?
3. How would the Riemann sum look if you had to approximate the area under the curve for finite $n$?
4. Can you extend the Riemann sum approach to other types of integrals like improper integrals?
5. What would change in the Riemann sum if the function being integrated had a discontinuity?

Tip: Always double-check your limits of integration when setting up Riemann sums, as it’s easy to mix up bounds and intervals.