To solve the integral [tex]\(\int_2^{+\infty} \frac{k}{x^2} \, dx\)[/tex], let's go through the steps methodically.
1. Set up the integral:
[tex]\[
\int_2^{+\infty} \frac{k}{x^2} \, dx
\][/tex]
2. Notice that [tex]\(k\)[/tex] is a constant and can be factored out of the integral:
[tex]\[
k \int_2^{+\infty} \frac{1}{x^2} \, dx
\][/tex]
3. Integrate the function [tex]\( \frac{1}{x^2} \)[/tex]:
Recall the formula for the integral of [tex]\(x^{-n}\)[/tex]:
[tex]\[
\int x^{-n} \, dx = \frac{x^{-n + 1}}{-n + 1} + C
\][/tex]
In our case, the integrand is [tex]\(x^{-2}\)[/tex]. So:
[tex]\[
\int x^{-2} \, dx = \int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = \frac{x^{-2 + 1}}{-2 + 1} = \frac{x^{-1}}{-1} = -\frac{1}{x}
\][/tex]
4. Evaluate the definite integral:
The indefinite integral we found is:
[tex]\[
-\frac{1}{x}
\][/tex]
We need to evaluate this from [tex]\(x = 2\)[/tex] to [tex]\( x = \infty\)[/tex]. Therefore, the definite integral becomes:
[tex]\[
\left[ -\frac{1}{x} \right]_2^{+\infty}
\][/tex]
5. Apply the limits of integration:
First, evaluate at the upper limit, [tex]\(\infty\)[/tex]:
[tex]\[
\lim_{x \to \infty} -\frac{1}{x} = 0
\][/tex]
Then, evaluate at the lower limit, 2:
[tex]\[
-\frac{1}{2}
\][/tex]
So, the definite integral is:
[tex]\[
0 - \left( -\frac{1}{2} \right) = 0 + \frac{1}{2} = \frac{1}{2}
\][/tex]
6. Multiply by the constant [tex]\( k \)[/tex]:
[tex]\[
k \cdot \frac{1}{2} = \frac{k}{2}
\][/tex]
Therefore, the final answer is:
[tex]\[
\int_2^{+\infty} \frac{k}{x^2} \, dx = \frac{k}{2}
\][/tex]
