Question

Evaluate \( \int_{1}^{2} \frac{1}{x^2} \, dx \).

• A. \( \frac{1}{2} \)
• B. \( 1 \)
• C. \( \frac{3}{2} \)
• D. \( 2 \)

Hint:

For integrals of the form \( \int x^n \, dx \), the antiderivative is \( \frac{x^{n+1}}{n+1} \) (for \( n \neq -1 \)); for \( n = -2 \), it’s \( -\frac{1}{x} \).

Answer 1

The Correct Option is A

To evaluate \( \int_{1}^{2} \frac{1}{x^2} \, dx = \int_{1}^{2} x^{-2} \, dx \), find the antiderivative: \[ \int x^{-2} \, dx = \int x^{-2} \, dx = \frac{x^{-1}}{-1} = -\frac{1}{x} + C \] Apply the definite integral from 1 to 2: \[ \left[ -\frac{1}{x} \right]_{1}^{2} = -\frac{1}{2} - \left(-\frac{1}{1}\right) = -\frac{1}{2} + 1 = \frac{1}{2} \] Thus, the value of the integral is: \[ \boxed{\frac{1}{2}} \]