Key Concepts and Formulas

• Area under a curve: The area bounded by the curve $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$ is given by $\int_a^b |f(x)| dx$. If $f(x) \ge 0$ on $[a,b]$, then the area is simply $\int_a^b f(x) dx$.

• Properties of Definite Integrals: $\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx$ and $\int_{-a}^a f(x) dx = 2\int_0^a f(x) dx$ if $f(x)$ is an even function.

• Even and Odd Functions: A function $f(x)$ is even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$.

Step-by-Step Solution

Step 1: Analyze the given region

We are given the region A = {(x, y) : 0 $\le$ y $\le$x |x| + 1 and $-$1 $\le$ x $\le$1}. We need to find the area of this region. The area is bounded by the x-axis (y=0), the curve y = x|x| + 1, and the vertical lines x = -1 and x = 1.

Step 2: Express the area as a definite integral

The area of the region A can be expressed as the definite integral: $\text{Area} = \int_{-1}^{1} (x|x| + 1) dx$

Step 3: Simplify the integrand

We can rewrite $x|x|$ as a piecewise function: $x|x| = \begin{cases} x^2, & \text{if } x \ge 0 \\ -x^2, & \text{if } x < 0 \end{cases}$ So, $x|x|$ is an odd function because $(-x)|-x| = -x|x|$. The function $y = x|x| + 1$ is the sum of an odd function ($x|x|$) and an even function (1).

Step 4: Evaluate the definite integral using properties of even and odd functions

We can split the integral into two parts: $\text{Area} = \int_{-1}^{1} x|x| dx + \int_{-1}^{1} 1 dx$ Since $x|x|$ is an odd function, its integral from -1 to 1 is 0: $\int_{-1}^{1} x|x| dx = 0$ The integral of 1 from -1 to 1 is: $\int_{-1}^{1} 1 dx = [x]_{-1}^{1} = 1 - (-1) = 2$ Therefore, the area is: $\text{Area} = 0 + 2 = 2$

Step 5: Re-evaluate the integral to arrive at the correct answer

The previous calculation is incorrect because we are considering the area between the x-axis and the curve $y=x|x|+1$. Since $y=x|x|+1$ is always positive in the interval [-1, 1], the area is given by $\int_{-1}^1 (x|x|+1) dx$.

Instead of using the property of even and odd functions directly, we can split the integral at x = 0 since the expression for |x| changes at x = 0: $\text{Area} = \int_{-1}^{0} (-x^2 + 1) dx + \int_{0}^{1} (x^2 + 1) dx$ $\text{Area} = \left[ -\frac{x^3}{3} + x \right]_{-1}^{0} + \left[ \frac{x^3}{3} + x \right]_{0}^{1}$ $\text{Area} = \left[ (0) - \left( -\frac{(-1)^3}{3} + (-1) \right) \right] + \left[ \left( \frac{(1)^3}{3} + (1) \right) - (0) \right]$ $\text{Area} = \left[ 0 - \left( \frac{1}{3} - 1 \right) \right] + \left[ \left( \frac{1}{3} + 1 \right) - 0 \right]$ $\text{Area} = \left[ 0 - \left( -\frac{2}{3} \right) \right] + \left[ \frac{4}{3} - 0 \right]$ $\text{Area} = \frac{2}{3} + \frac{4}{3} = \frac{6}{3} = 2$

There must be an error in the problem statement or the provided answer. We have calculated the area correctly as 2. However, we are told the correct answer is $\frac{2}{3}$.

Let's analyze the problem again. The region is bounded by $0 \le y \le x|x| + 1$ and $-1 \le x \le 1$. We found the correct integral setup as $\text{Area} = \int_{-1}^{1} (x|x| + 1) dx = \int_{-1}^{0} (-x^2 + 1) dx + \int_{0}^{1} (x^2 + 1) dx$

If the problem was $0 \le y \le |x|x + 1$ and $-1 \le x \le 0$, we would have $\int_{-1}^0 (-x^2 + 1) dx = [-\frac{x^3}{3} + x]_{-1}^0 = 0 - (\frac{1}{3} - 1) = \frac{2}{3}$ If the problem was $0 \le y \le x^2$, we would have $\int_{-1}^1 x^2 dx = [\frac{x^3}{3}]_{-1}^1 = \frac{1}{3} - (-\frac{1}{3}) = \frac{2}{3}$

The correct area calculation gives 2. The answer provided is 2/3. Let's assume the question meant to ask for the area between $y = x|x|+1$ and $y=1$ for $-1 \le x \le 1$. Then the area would be $\int_{-1}^1 (x|x|+1 - 1) dx = \int_{-1}^1 x|x| dx = 0$

Let's assume the region is defined by $0 \le y \le 1 - x|x|$ and $-1 \le x \le 1$. $\text{Area} = \int_{-1}^1 (1 - x|x|) dx = \int_{-1}^0 (1+x^2) dx + \int_0^1 (1 - x^2) dx$ $= [x + \frac{x^3}{3}]_{-1}^0 + [x - \frac{x^3}{3}]_0^1 = 0 - (-1 - \frac{1}{3}) + (1 - \frac{1}{3}) - 0 = 1 + \frac{1}{3} + 1 - \frac{1}{3} = 2$

If the question was $y \le x|x| + 1$ and $y \ge 1$ for $-1 \le x \le 1$, then the area would be 0.

The area of the region is calculated as 2. But the correct answer is given as $\frac{2}{3}$. This is impossible. There must be an error in the question or the provided answer. I will assume that the question is correct and the answer provided is incorrect.

Step 6: Conclusion

The correct area of the region A = {(x, y) : 0 $\le$ y $\le$x |x| + 1 and $-$1 $\le$ x $\le$1} is 2.

Common Mistakes & Tips

• Remember to consider the absolute value when finding the area under a curve.
• Splitting the integral at x=0 can be helpful when dealing with absolute values.
• Always double-check your calculations to avoid errors.
• When dealing with even and odd functions, remember that the integral of an odd function over a symmetric interval is zero.

Summary

We found the area of the region bounded by $y = x|x| + 1$, the x-axis, and the lines $x = -1$ and $x = 1$. We split the integral into two parts to account for the absolute value and used the properties of definite integrals. The area was found to be 2. The provided answer is incorrect.

Final Answer

The final answer is \boxed{2}.