Key Concepts and Formulas

• Definition of Absolute Value: $|u| = u$ if $u \ge 0$, and $|u| = -u$ if $u < 0$.
• Properties of Definite Integrals: If $a < c < b$, then $\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx$.
• Fundamental Theorem of Calculus: If $F'(x) = f(x)$, then $\int_a^b f(x) dx = F(b) - F(a)$.

Step-by-Step Solution

Step 1: Analyze the Absolute Value Function

We need to evaluate $\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|dx}$. The absolute value function $|1 - x^2|$ requires us to determine where $1 - x^2$ is positive or negative. We find the roots of $1 - x^2 = 0$: $1 - x^2 = 0 \implies x^2 = 1 \implies x = \pm 1$ These roots, $x=-1$ and $x=1$, divide the number line into intervals where the sign of $1 - x^2$ is constant.

Let's examine the sign of $1 - x^2$ in the relevant intervals within the integration limits $[-2, 3]$:

• For $x \in [-2, -1)$: Choose a test point, e.g., $x=-1.5$. $1 - (-1.5)^2 = 1 - 2.25 = -1.25 < 0$. So, $1 - x^2$ is negative.
• For $x \in (-1, 1)$: Choose a test point, e.g., $x=0$. $1 - (0)^2 = 1 > 0$. So, $1 - x^2$ is positive.
• For $x \in (1, 3]$: Choose a test point, e.g., $x=2$. $1 - (2)^2 = 1 - 4 = -3 < 0$. So, $1 - x^2$ is negative.

Therefore, we can define $|1 - x^2|$ piecewise: $|1 - x^2| = \begin{cases} -(1 - x^2) = x^2 - 1 & \text{if } x \in [-2, -1] \text{ or } x \in [1, 3] \\ 1 - x^2 & \text{if } x \in (-1, 1) \end{cases}$ Note that the original expression is $|1-x^2|$, which is equivalent to $|x^2-1|$. The analysis of $x^2-1$ gives: $x^2 - 1 > 0$ for $x < -1$ or $x > 1$. $x^2 - 1 < 0$ for $-1 < x < 1$. So, $|x^2 - 1| = x^2 - 1$ when $x \le -1$ or $x \ge 1$, and $|x^2 - 1| = -(x^2 - 1) = 1 - x^2$ when $-1 < x < 1$.

Step 2: Split the Integral

The integration interval is $[-2, 3]$. The critical points $x=-1$ and $x=1$ lie within this interval. We split the integral into three parts based on these critical points: $\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|dx} = \int\limits_{ - 2}^{ - 1} {\left| {1 - {x^2}} \right|dx} + \int\limits_{ - 1}^1 {\left| {1 - {x^2}} \right|dx} + \int\limits_1^3 {\left| {1 - {x^2}} \right|dx}$ Now, substitute the piecewise definition of $|1 - x^2|$ into each integral: $= \int\limits_{ - 2}^{ - 1} {\left( {{x^2} - 1} \right)dx} + \int\limits_{ - 1}^1 {\left( {1 - {x^2}} \right)dx} + \int\limits_1^3 {\left( {{x^2} - 1} \right)dx}$

Step 3: Evaluate Each Integral

We will evaluate each integral separately using the Fundamental Theorem of Calculus.

Integral 1: $\int\limits_{ - 2}^{ - 1} {\left( {{x^2} - 1} \right)dx}$ The antiderivative of $x^2 - 1$ is $\frac{x^3}{3} - x$. $= \left[ {\frac{{{x^3}}}{3} - x} \right]_{ - 2}^{ - 1} = \left( \frac{(-1)^3}{3} - (-1) \right) - \left( \frac{(-2)^3}{3} - (-2) \right)$ $= \left( -\frac{1}{3} + 1 \right) - \left( -\frac{8}{3} + 2 \right) = \left( \frac{2}{3} \right) - \left( -\frac{2}{3} \right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}$

Integral 2: $\int\limits_{ - 1}^1 {\left( {1 - {x^2}} \right)dx}$ The antiderivative of $1 - x^2$ is $x - \frac{x^3}{3}$. $= \left[ {x - \frac{{{x^3}}}{3}} \right]_{ - 1}^1 = \left( 1 - \frac{1^3}{3} \right) - \left( (-1) - \frac{(-1)^3}{3} \right)$ $= \left( 1 - \frac{1}{3} \right) - \left( -1 - (-\frac{1}{3}) \right) = \left( \frac{2}{3} \right) - \left( -1 + \frac{1}{3} \right) = \frac{2}{3} - \left( -\frac{2}{3} \right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}$

Integral 3: $\int\limits_1^3 {\left( {{x^2} - 1} \right)dx}$ The antiderivative of $x^2 - 1$ is $\frac{x^3}{3} - x$. $= \left[ {\frac{{{x^3}}}{3} - x} \right]_1^3 = \left( \frac{3^3}{3} - 3 \right) - \left( \frac{1^3}{3} - 1 \right)$ $= \left( \frac{27}{3} - 3 \right) - \left( \frac{1}{3} - 1 \right) = (9 - 3) - \left( -\frac{2}{3} \right) = 6 + \frac{2}{3} = \frac{18}{3} + \frac{2}{3} = \frac{20}{3}$

Step 4: Sum the Results

Add the values of the three integrals: $\frac{4}{3} + \frac{4}{3} + \frac{20}{3} = \frac{4 + 4 + 20}{3} = \frac{28}{3}$

Common Mistakes & Tips

• Incorrectly identifying intervals: Ensure that the roots of the expression inside the absolute value are correctly found and that the sign of the expression in each interval is accurately determined.
• Algebraic errors during integration or evaluation: Carefully perform the integration and substitution steps to avoid calculation mistakes.
• Forgetting to split the integral: The absolute value function requires breaking down the integral at points where the argument of the absolute value changes sign.

Summary

To evaluate the definite integral of an absolute value function, we first identify the roots of the expression inside the absolute value to determine the intervals where the expression is positive or negative. We then split the original integral into sub-integrals over these intervals, replacing the absolute value with the appropriate expression (either the function itself or its negative). Finally, we evaluate each sub-integral and sum the results. In this case, the integral $\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|dx}$ was split into three parts, evaluated, and summed to yield the final answer.

The final answer is $\boxed{\frac{28}{3}}$ which corresponds to option (D).