Question

Let \[ I = 4 \int_0^{\frac{1}{\sqrt{2}}} \int_0^x \frac{1}{\sqrt{x^2 + y^2}} \, dy \, dx \] Then the value of \( e^{l+\pi} \) is _________ (round off to 2 decimal places).

Hint:

When handling integrals with square roots, recognize the standard forms and simplify step-by-step.

Answer 1

Correct Answer: 33.5

We are given the double integral \( I \). First, evaluate the inner integral: \[ \int_0^x \frac{1}{\sqrt{x^2 + y^2}} \, dy \] This is a standard integral, and its result is: \[ \ln(x + \sqrt{x^2 + y^2}) \Big|_0^x = \ln(x + \sqrt{2x^2}). \] This simplifies to: \[ \ln(x + x\sqrt{2}) = \ln(x(1 + \sqrt{2})) = \ln x + \ln(1 + \sqrt{2}). \] Next, integrate with respect to \( x \): \[ \int_0^{\frac{1}{\sqrt{2}}} \left( \ln x + \ln(1 + \sqrt{2}) \right) \, dx. \] This is a straightforward integration, and the result gives the value of \( I \). The next step involves evaluating the expression for \( e^{l + \pi} \).