The problem asks us to evaluate the definite integral $\int_1^2 (\frac{1}{x^2} - \frac{1}{x^3}) \, dx$ using the properties of integrals, given that $\int_1^2 \frac{1}{x^2} \, dx = \frac{1}{2}$ and $\int_1^2 \frac{1}{x^3} \, dx = \frac{3}{8}$.

1. Problem Description

The problem asks us to evaluate the definite integral $\int_1^2 (\frac{1}{x^2} - \frac{1}{x^3}) \, dx$ using the properties of integrals, given that $\int_1^2 \frac{1}{x^2} \, dx = \frac{1}{2}$ and $\int_1^2 \frac{1}{x^3} \, dx = \frac{3}{8}$.

2. Solution Steps

We can use the property of linearity of integrals, which states that the integral of a difference of two functions is the difference of their integrals:

$$\int_a^b [f(x) - g(x)] \, dx = \int_a^b f(x) \, dx - \int_a^b g(x) \, dx$$

In our case, $f(x) = \frac{1}{x^2}$ and $g(x) = \frac{1}{x^3}$, $a = 1$, and $b = 2$. Therefore,

$$\int_1^2 \left(\frac{1}{x^2} - \frac{1}{x^3}\right) \, dx = \int_1^2 \frac{1}{x^2} \, dx - \int_1^2 \frac{1}{x^3} \, dx$$

We are given that $\int_1^2 \frac{1}{x^2} \, dx = \frac{1}{2}$ and $\int_1^2 \frac{1}{x^3} \, dx = \frac{3}{8}$.

Substituting these values into the equation, we get:

$$\int_1^2 \left(\frac{1}{x^2} - \frac{1}{x^3}\right) \, dx = \frac{1}{2} - \frac{3}{8}$$

To subtract the fractions, we need a common denominator, which is

8. So, we rewrite $\frac{1}{2}$ as $\frac{4}{8}$:

$$\int_1^2 \left(\frac{1}{x^2} - \frac{1}{x^3}\right) \, dx = \frac{4}{8} - \frac{3}{8} = \frac{4-3}{8} = \frac{1}{8}$$

3. Final Answer

The value of the definite integral is $\frac{1}{8}$.