Key Concepts and Formulas

• Property of Definite Integrals for Symmetric Limits: If $f(x)$ is an integrable function on $[-a, a]$, then $\int_{-a}^a f(x)dx = \int_0^a [f(x) + f(-x)]dx$.
• Odd and Even Functions:
  • A function $f(x)$ is odd if $f(-x) = -f(x)$. For an odd function, $\int_{-a}^a f(x)dx = 0$.
  • A function $f(x)$ is even if $f(-x) = f(x)$. For an even function, $\int_{-a}^a f(x)dx = 2\int_0^a f(x)dx$.
• Definition of Absolute Value: $|x| = x$ if $x \ge 0$, and $|x| = -x$ if $x < 0$.

Step-by-Step Solution

Let the given integral be $I$. $I = \int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {({e^{x|x|}} + 1)}}dx}$

Step 1: Analyze the integrand and the interval of integration. The interval of integration is $[-2, 2]$, which is symmetric about $0$. This suggests using properties of definite integrals for symmetric intervals. The integrand is $f(x) = {{{|{x^3} + x|} \over {({e^{x|x|}} + 1)}}}$.

Step 2: Simplify the absolute value term. Consider the term $|x^3 + x|$. We can factor out $x$: $|x^3 + x| = |x(x^2 + 1)|$. Since $x^2 + 1$ is always positive, $|x(x^2 + 1)| = |x|(x^2 + 1)$.

Step 3: Simplify the exponential term. Consider the term $e^{x|x|}$. If $x \ge 0$, then $|x| = x$, so $e^{x|x|} = e^{x \cdot x} = e^{x^2}$. If $x < 0$, then $|x| = -x$, so $e^{x|x|} = e^{x \cdot (-x)} = e^{-x^2}$.

Step 4: Rewrite the integrand using the simplified terms. The integrand becomes: $f(x) = \frac{|x|(x^2 + 1)}{e^{x|x|} + 1}$

Step 5: Apply the property of definite integrals for symmetric limits. We use the property $\int_{-a}^a f(x)dx = \int_0^a [f(x) + f(-x)]dx$. In this case, $a=2$. Let's find $f(-x)$: $f(-x) = \frac{|-x|((-x)^2 + 1)}{e^{(-x)|-x|} + 1}$ Since $|-x| = |x|$ and $(-x)^2 = x^2$, we have: $f(-x) = \frac{|x|(x^2 + 1)}{e^{-x|x|} + 1}$

Now, let's find $f(x) + f(-x)$: $f(x) + f(-x) = \frac{|x|(x^2 + 1)}{e^{x|x|} + 1} + \frac{|x|(x^2 + 1)}{e^{-x|x|} + 1}$ Factor out the common term $\frac{|x|(x^2 + 1)}{1}$: $f(x) + f(-x) = |x|(x^2 + 1) \left( \frac{1}{e^{x|x|} + 1} + \frac{1}{e^{-x|x|} + 1} \right)$ To simplify the term in the parenthesis, find a common denominator: $\frac{1}{e^{x|x|} + 1} + \frac{1}{e^{-x|x|} + 1} = \frac{e^{-x|x|} + 1 + e^{x|x|} + 1}{(e^{x|x|} + 1)(e^{-x|x|} + 1)}$ $= \frac{e^{x|x|} + e^{-x|x|} + 2}{e^{x|x|}e^{-x|x|} + e^{x|x|} + e^{-x|x|} + 1}$ Since $e^{x|x|}e^{-x|x|} = e^0 = 1$, the denominator becomes $1 + e^{x|x|} + e^{-x|x|} + 1 = e^{x|x|} + e^{-x|x|} + 2$. So, the expression in the parenthesis simplifies to: $\frac{e^{x|x|} + e^{-x|x|} + 2}{e^{x|x|} + e^{-x|x|} + 2} = 1$

Therefore, $f(x) + f(-x) = |x|(x^2 + 1) \cdot 1 = |x|(x^2 + 1)$.

Step 6: Evaluate the integral from 0 to 2. Using the property, the integral becomes: $I = \int_0^2 [f(x) + f(-x)]dx = \int_0^2 |x|(x^2 + 1)dx$ Since the integration is from $0$ to $2$, $x \ge 0$, so $|x| = x$. $I = \int_0^2 x(x^2 + 1)dx$ $I = \int_0^2 (x^3 + x)dx$

Step 7: Compute the definite integral. $I = \left[ \frac{x^4}{4} + \frac{x^2}{2} \right]_0^2$ $I = \left( \frac{2^4}{4} + \frac{2^2}{2} \right) - \left( \frac{0^4}{4} + \frac{0^2}{2} \right)$ $I = \left( \frac{16}{4} + \frac{4}{2} \right) - (0)$ $I = (4 + 2)$ $I = 6$

Common Mistakes & Tips

• Incorrectly handling the absolute value: Ensure that $|x^3 + x|$ is correctly simplified as $|x|(x^2+1)$, recognizing that $x^2+1$ is always positive.
• Errors in simplifying the exponential term: Be careful with the sign of the exponent in $e^{x|x|}$ when substituting $-x$ for $x$.
• Algebraic errors in combining fractions: Double-check the simplification of the sum of the two fractions in the parenthesis.

Summary

The problem involves evaluating a definite integral over a symmetric interval. By analyzing the integrand and applying the property $\int_{-a}^a f(x)dx = \int_0^a [f(x) + f(-x)]dx$, we simplified the problem significantly. The key steps involved simplifying the absolute value and exponential terms, and then combining $f(x)$ and $f(-x)$. The resulting integral was straightforward to compute.

The final answer is $\boxed{6}$.