Question

The value of \( \int_{0}^{1} \frac{1}{3x - 4} \, dx \) is :

• A. \( \frac{1}{3} \log 4 \)
• B. \( -\frac{1}{3} \log 4 \)
• C. \( \log (-4) \)
• D. \( \log 4 \)

Hint:

Always use absolute value signs within the logarithm argument during integration, because \( \log(x) \) is undefined for negative real numbers. This ensures that terms like \( |-4| \) correctly simplify to \( 4 \).

Answer 1

The Correct Option is B

Concept: The standard integration formula for a linear fractional form is \( \int \frac{1}{ax + b} \, dx = \frac{1}{a} \log |ax + b| \). When dealing with definite integration, evaluate this anti-derivative at the upper limit and subtract its value at the lower limit.

Step 1: Compute the indefinite integral. Let us find the antiderivative of the function: \[ \int \frac{1}{3x - 4} \, dx = \frac{1}{3} \log |3x - 4| \]

Step 2: Apply the limits of integration from \( 0 \) to \( 1 \). Using the Fundamental Theorem of Calculus: \[ I = \left[ \frac{1}{3} \log |3x - 4| \right]_{0}^{1} \] Substitute the upper limit \( x = 1 \): \[ I_{\text{upper}} = \frac{1}{3} \log |3(1) - 4| = \frac{1}{3} \log |3 - 4| = \frac{1}{3} \log |-1| = \frac{1}{3} \log(1) = 0 \] Substitute the lower limit \( x = 0 \): \[ I_{\text{lower}} = \frac{1}{3} \log |3(0) - 4| = \frac{1}{3} \log |0 - 4| = \frac{1}{3} \log |-4| = \frac{1}{3} \log 4 \]

Step 3: Subtract lower limit value from upper limit value. \[ I = I_{\text{upper}} - I_{\text{lower}} = 0 - \frac{1}{3} \log 4 = -\frac{1}{3} \log 4 \]