Question

use the fact that \\(\int\limits_{a}^{b} x \\, dx = \frac{b^2}{2} - \frac{a^2}{2}\\), where \\(a < b\\), to evaluate the integral \\(\int\limits_{1.5}^{2.5} x \\, dx\\). the value of the integral \\(\int\limits_{1.5}^{2.5} x \\, dx\\) is \\(\square\\).

Explanation:

Step1: Identify a and b

In the integral $\int_{1.5}^{2.5} x \, dx$, we have $a = 1.5$ and $b = 2.5$.

Step2: Apply the formula

Using the formula $\int_{a}^{b} x \, dx = \frac{b^2}{2} - \frac{a^2}{2}$, substitute $a = 1.5$ and $b = 2.5$.
First, calculate $b^2$ and $a^2$:
$b^2 = (2.5)^2 = 6.25$
$a^2 = (1.5)^2 = 2.25$
Then, substitute into the formula:
$\frac{6.25}{2} - \frac{2.25}{2} = \frac{6.25 - 2.25}{2} = \frac{4}{2} = 2$

Answer:

2