$$ \int_0^{58}\!\int_0^{\frac{41}{58}x}\!e^{-x^2}\,\mathrm{dy}\,\mathrm{d}x $$

To evaluate the double integral 

$$
\int_0^{58}\int_0^{\frac{41}{58}x} e^{-x^2} \, \mathrm{dy} \, \mathrm{d}x,
$$

we can start by evaluating the inner integral with respect to $ y $. Since $ e^{-x^2} $ is independent of $ y $, we can pull it out of the inner integral:

\[
\int_0^{\frac{41}{58}x} e^{-x^2} \, \mathrm{dy} = e^{-x^2} \int_0^{\frac{41}{58}x} \mathrm{dy}.
\]

The integral of $ 1 $ with respect to $ y $ is simply $ y $, so we get:

\[
\int_0^{\frac{41}{58}x} \mathrm{dy} = y \Big|_0^{\frac{41}{58}x} = \frac{41}{58}x.
\]

So the inner integral simplifies to:

\[
e^{-x^2} \cdot \frac{41}{58}x.
\]

Now, substitute this result back into the outer integral:

\[
\int_0^{58} \frac{41}{58}x e^{-x^2} \, \mathrm{d}x.
\]

We can pull the constant $ \frac{41}{58} $ out of the integral:

\[
\frac{41}{58} \int_0^{58} x e^{-x^2} \, \mathrm{d}x.
\]

To evaluate this integral, we use the substitution $ u = x^2 $, hence $ du = 2x \, \mathrm{d}x $, and $ x \, \mathrm{d}x = \frac{1}{2} du $. The limits of integration for $ u $ when $ x = 0 $ and $ x = 58 $ are $ u = 0 $ and $ u = 58^2 = 3364 $, respectively. The integral then becomes:

\[
\frac{41}{58} \cdot \frac{1}{2} \int_0^{3364} e^{-u} \, \mathrm{d}u.
\]

The integral of $ e^{-u} $ with respect to $ u $ is $ -e^{-u} $, so we have:

\[
\frac{41}{116} \left[-e^{-u} \right]_0^{3364} = \frac{41}{116} \left[-e^{-3364} + e^0\right].
\]

Simplifying, we get:

\[
\frac{41}{116} \left[1 - e^{-3364} \right].
\]

Given that $ e^{-3364} $ is an extremely small number, we can approximate it as 0, leaving us with:

\[
\frac{41}{116} \approx 0.3534.
\]

So the value of the double integral is approximately $ 0.3534 $.

Do you want more details or have any questions?

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**Related Questions:**

1. What is the significance of the substitution method in evaluating integrals?
2. How do you handle double integrals with variable limits of integration?
3. Why can $ e^{-3364} $ be approximated as 0?
4. What are some common techniques for solving double integrals?
5. How does the order of integration affect the evaluation of a double integral?
6. Can this integral be solved using numerical methods instead of analytical methods?
7. What are some applications of double integrals in physics and engineering?
8. How would the integral change if the limits were different?

**Tip:** When dealing with integrals involving exponential functions, substitutions often simplify the process significantly.

Learn how to evaluate the double integral ∫₀⁵⁸ ∫₀^(⁴¹/₅₈ₓ) e^(-x²) dy dx using the substitution method. Detailed step-by-step solution with key formulas and methods explained. Suitable for advanced college-level mathematics.

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