Question:

Given that the integral \( I = \int_{x}^{i} \log t \, dt = \frac{1}{4} \), find the value of \( x \).

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When solving integrals involving logarithms, use the standard integration formula for \( \log t \) and apply the limits carefully.

Updated On: Feb 14, 2025

\( \sqrt{e} \)
\( e \)
\( 1 \)
\( 2 \)

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The Correct Option is A

Solution and Explanation

Step 1: Solve the integral.

We are given the integral: \[ I = \int_{x}^{i} \log t \, dt \] The integral of \( \log t \) is: \[ \int \log t \, dt = t \log t - t + C \]

Step 2: Apply the limits of integration. Now, apply the limits \( x \) to \( i \): \[ I = \left[ t \log t - t \right]_{x}^{i} = \left( i \log i - i \right) - \left( x \log x - x \right) \] We are given that \( I = \frac{1}{4} \), so: \[ i \log i - i - (x \log x - x) = \frac{1}{4} \]

Step 3: Simplify the equation. Using \( \log i = 1 \) (since \( \log e = 1 \)) and simplifying the equation: \[ i - x \log x + x = \frac{1}{4} \]

Thus, we find that the value of \( x \) is \( \sqrt{e} \).

Therefore, the correct answer is \( \sqrt{e} \).

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Consider the following code:

int a;
int arr[] = {30, 50, 10};
int *ptr = arr[10] + 1;
a = *ptr;
(*ptr)++;
ptr = ptr + 1;
printf("%d", a + arr[1] + *ptr);

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What is the output of the following C code?

  void foo(int *p, int x) {
      *p = x;
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      z = a;  // Incorrect: Should be z = a;
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Issue: The statement z = a; is invalid because a is an integer, and z is a pointer.

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