Question

compute the the exact value of the given integral.\\(\int_{4}^{6} \frac{1}{x^2} dx = \\)

Explanation:

Step1: Rewrite integrand

Rewrite $\frac{1}{x^2}$ as $x^{-2}$.

Step2: Find antiderivative

Antiderivative of $x^{-2}$ is $\frac{x^{-1}}{-1} = -\frac{1}{x}$.

Step3: Apply Fundamental Theorem

Evaluate $-\frac{1}{x}$ from 4 to 6: $(-\frac{1}{6}) - (-\frac{1}{4})$.

Step4: Calculate result

Simplify: $-\frac{1}{6} + \frac{1}{4} = \frac{-2 + 3}{12} = \frac{1}{12}$.

Answer:

$\frac{1}{12}$