1 markMCQSection A

If $\int_{0}^{40} \frac{dx}{2x + 1} = \log k$ , then the value of k is :

Q: If $\int_{0}^{40} \frac{dx}{2x + 1} = \log k$, then the value of k is :

The correct option is **B) 9**. **Working:** - $\int \frac{dx}{2x+1} = \frac{1}{2}\log(2x+1)$. - Evaluate from 0 to 40: $\frac{1}{2}[\log(2 \cdot 40+1) - \log(2 \cdot 0+1)] = \frac{1}{2}[\log 81 - \log 1]$. - $= \frac{1}{2} \log 81 = \log 81^{1/2} = \log 9$. - Comparing with $\log k$ gives $k = 9$.

Calculus (Integration)

Definite Integration

Options

(A)

3

(B)

9

(C)

\frac{9}{2}

(D)

\frac{3}{2}

Official Answer

The correct option is B) 9.

Working:

$\int \frac{dx}{2x+1} = \frac{1}{2}\log(2x+1)$ .
Evaluate from 0 to 40: $\frac{1}{2}[\log(2 \cdot 40+1) - \log(2 \cdot 0+1)] = \frac{1}{2}[\log 81 - \log 1]$ .
$= \frac{1}{2} \log 81 = \log 81^{1/2} = \log 9$ .
Comparing with $\log k$ gives $k = 9$ .

definite integralintegral of 1/(2x+1)(1/2) log(2x+1)log 81log 9k = 9square root inside log

Marking Scheme

11 mark: correct antiderivative $\frac{1}{2}\log(2x+1)$ , evaluation to $\frac{1}{2}\log 81 = \log 9$ , and $k = 9$ (option B).

Hint

Integrate to $\frac{1}{2}\log(2x+1)$ ; then $\frac{1}{2}\log 81 = \log 81^{1/2} = \log 9$ .

Quick Oral Answer

The integral is $\frac{1}{2}\log(2x+1)$ ; from 0 to 40 that is $\frac{1}{2}\log 81 = \log 9$ , so k equals 9.

Analysis & Explanation

This tests evaluation of a definite integral of the form $\int \frac{dx}{ax+b}$ and log manipulation.

Concept:

$\int \frac{dx}{2x+1} = \frac{1}{2}\log(2x+1)$ because of the inner-derivative factor 2.
The coefficient 1/2 becomes a power inside the log: $\frac{1}{2}\log 81 = \log 9$ .

Why B is correct:

$\frac{1}{2}[\log 81 - \log 1] = \frac{1}{2}\log 81 = \log 9$ , so $k = 9$ .

Why the distractors are wrong:

A (3): from $\sqrt{9} = 3$ , mistaking $\sqrt{81} = 9$ for a further root.
C ( $\frac{9}{2}$ ): from forgetting to fold the 1/2 into the log and instead multiplying 9 by 1/2 wrongly, or halving 9.
D ( $\frac{3}{2}$ ): a compounded version of the same 1/2 and root errors.

Common Mistakes

1Omitting the factor $\frac{1}{2}$ from the antiderivative, giving $\log 81$ and $k = 81$ .
2Converting $\frac{1}{2}\log 81$ wrongly to $\log(81/2)$ instead of $\log 81^{1/2}$ .
3Miscalculating the upper limit: $2(40)+1 = 81$ , not 80 or 41.

Interesting Facts

The rule $\int \frac{dx}{ax+b} = \frac{1}{a}\log|ax+b|$ is one of the most frequently used standard integrals in applied calculus.

Since $81 = 3^4$ , $\sqrt{81} = 9 = 3^2$ — the powers of 3 make this integral evaluate to a clean logarithm.

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Frequently Asked Questions

Where does the $\frac{1}{2}$ factor come from?

Because the derivative of the inner expression 2x+1 is 2, integrating $\frac{1}{2x+1}$ introduces a 1/2, giving $\frac{1}{2}\log(2x+1)$ .

How does $\frac{1}{2}\log 81$ become $\log 9$ ?

A coefficient in front of a log becomes a power inside it: $\frac{1}{2}\log 81 = \log 81^{1/2} = \log \sqrt{81} = \log 9$ .