Question:

The value of \( \int_{-5}^{-1} \frac{1}{x} \, dx \) is equal to :

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Always remember the absolute value: \( \int \frac{1}{x} dx = \log|x| \). Without it, you might incorrectly try to evaluate \( \log(-1) \).
Property: \( \log(1/a) = -\log a \).
Updated On: Sep 10, 2026
  • \( -\log 5 \)
  • \( \log 5 \)
  • \( \log(-5) \)
  • \( 0 \)
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The Correct Option is A

Solution and Explanation

Concept:
• The integral of \( \frac{1}{x} \) is \( \log |x| \).
• The Fundamental Theorem of Calculus states \( \int_a^b f(x) \, dx = F(b) - F(a) \).
• Always use the absolute value inside the logarithm for real-valued integrals of \( 1/x \).

Step 1:
Apply the integration rule for \( 1/x \)
Let the integral be \( I = \int_{-5}^{-1} \frac{1}{x} \, dx \). The antiderivative of \( \frac{1}{x} \) is \( \log |x| \). \[ I = [\log |x|]_{-5}^{-1} \]

Step 2:
Substitute the upper and lower limits
\[ I = \log |-1| - \log |-5| \] Since \( |-1| = 1 \) and \( |-5| = 5 \): \[ I = \log 1 - \log 5 \]

Step 3:
Simplify the final value
We know that \( \log 1 = 0 \) for any base. \[ I = 0 - \log 5 \] \[ I = -\log 5 \] This matches option (A).
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