Math Problem Statement

int 0 ^ 2 int 0 ^ x^ 2 e^ y x dxdy

Solution

You have written a double integral expression, which I assume is meant to evaluate the following:

020x2eydxdy\int_0^2 \int_0^{x^2} e^y \, dx \, dy

Here's how we would approach solving this:

Step 1: Understand the limits and function

The given integral is a double integral, which means we will integrate the function eye^y over a two-dimensional region. The limits of integration are:

  • For yy, it goes from 0 to x2x^2.
  • For xx, it goes from 0 to 2.

Thus, the order of integration is: first, with respect to xx, and then with respect to yy.

Step 2: Solve the inner integral (with respect to xx)

We first integrate with respect to xx. The inner integral is:

02(0x2eydy)dx\int_0^2 \left( \int_0^{x^2} e^y \, dy \right) dx

Since the inner integral is with respect to yy, and the function eye^y is independent of xx, we can integrate eye^y:

0x2eydy=[ey]0x2=ex2e0=ex21\int_0^{x^2} e^y \, dy = \left[ e^y \right]_0^{x^2} = e^{x^2} - e^0 = e^{x^2} - 1

Step 3: Solve the outer integral (with respect to xx)

Now we substitute this result back into the outer integral:

02(ex21)dx\int_0^2 \left( e^{x^2} - 1 \right) dx

This is now a single integral in terms of xx. It involves ex2e^{x^2}, which does not have a simple antiderivative in terms of elementary functions. However, we can split the integral as follows:

02(ex21)dx=02ex2dx021dx\int_0^2 \left( e^{x^2} - 1 \right) dx = \int_0^2 e^{x^2} dx - \int_0^2 1 \, dx

Step 4: Solve each part separately

  • The integral 021dx\int_0^2 1 \, dx is straightforward: 021dx=2\int_0^2 1 \, dx = 2

  • The integral 02ex2dx\int_0^2 e^{x^2} dx is a known function that is related to the error function (erf(x)\text{erf}(x)):

02ex2dx=π2erfi(2)\int_0^2 e^{x^2} dx = \frac{\sqrt{\pi}}{2} \text{erfi}(2)

Here, erfi(x)\text{erfi}(x) is the imaginary error function, but you would typically approximate it numerically.

Final Answer:

So, the double integral evaluates to:

020x2eydydx=(π2erfi(2)2)\int_0^2 \int_0^{x^2} e^y \, dy \, dx = \left( \frac{\sqrt{\pi}}{2} \text{erfi}(2) - 2 \right)

Would you like me to continue with numerical approximation or go into further details?

Here are 5 related questions for further exploration:

  1. How does changing the limits of integration affect the result of a double integral?
  2. What is the error function, and how is it related to integrals involving exponentials?
  3. What is the difference between erf(x)\text{erf}(x) and erfi(x)\text{erfi}(x)?
  4. Can we solve integrals involving functions like ex2e^{x^2} exactly, or are they usually approximated?
  5. How does changing the order of integration impact the result of a double integral?

Tip: When dealing with complex integrals, especially those involving functions like ex2e^{x^2}, knowing about special functions (like the error function) can be very helpful.

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Math Problem Analysis

Mathematical Concepts

Double Integral
Exponential Functions
Error Function
Change of Integration Order

Formulas

\int_0^2 \int_0^{x^2} e^y \, dy \, dx
e^y
erfi(x)

Theorems

Fubini's Theorem
Error Function Definition
Fundamental Theorem of Calculus

Suitable Grade Level

Undergraduate Calculus