The area of the region $$A = \{(x, y) : 0 \leq y \leq x|x| + 1$$ and $$-1 \leq x \leq 1\}$$ in sq. units, is:

We are asked to calculate the area enclosed by the set $$A=\{(x,y):0\le y\le x|x|+1\text{ and }-1\le x\le1\}.$$

First we analyse the upper boundary $$y=x|x|+1.$$ Because of the absolute value, this expression changes its form on either side of $$x=0.$$ For $$x\ge0$$ we have $$|x|=x,$$ so

$$y=x|x|+1=x\cdot x+1=x^{2}+1.$$

For $$x\le0$$ we have $$|x|=-x,$$ giving

$$y=x|x|+1=x(-x)+1=-x^{2}+1.$$

Hence the region is naturally split into two parts:

• On $$[-1,0],$$ the top curve is $$y=-x^{2}+1.$$

• On $$[0,1],$$ the top curve is $$y=x^{2}+1.$$

The lower boundary is the $$x$$-axis $$y=0$$ throughout. So the required area is the sum of two definite integrals:

$$\text{Area}=\int_{-1}^{0}\bigl((-x^{2}+1)-0\bigr)\,dx\;+\;\int_{0}^{1}\bigl((x^{2}+1)-0\bigr)\,dx.$$

We now evaluate each integral step by step.

For the first integral we integrate term by term:

$$\int_{-1}^{0}(-x^{2}+1)\,dx=\int_{-1}^{0}1\,dx-\int_{-1}^{0}x^{2}\,dx.$$

Using the power rule $$\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C,$$ we get

$$\int_{-1}^{0}1\,dx=\Bigl[x\Bigr]_{-1}^{0}=0-(-1)=1,$$

$$\int_{-1}^{0}x^{2}\,dx=\Bigl[\dfrac{x^{3}}{3}\Bigr]_{-1}^{0}=0-\Bigl(-\dfrac{1}{3}\Bigr)=\dfrac{1}{3}.$$

Subtracting,

$$\int_{-1}^{0}(-x^{2}+1)\,dx=1-\dfrac{1}{3}=\dfrac{2}{3}.$$

For the second integral we proceed similarly:

$$\int_{0}^{1}(x^{2}+1)\,dx=\int_{0}^{1}x^{2}\,dx+\int_{0}^{1}1\,dx.$$

Again applying the power rule,

$$\int_{0}^{1}x^{2}\,dx=\Bigl[\dfrac{x^{3}}{3}\Bigr]_{0}^{1}=\dfrac{1}{3}-0=\dfrac{1}{3},$$

$$\int_{0}^{1}1\,dx=\Bigl[x\Bigr]_{0}^{1}=1-0=1.$$

Adding,

$$\int_{0}^{1}(x^{2}+1)\,dx=\dfrac{1}{3}+1=\dfrac{4}{3}.$$

Now we add the two pieces to obtain the total area:

$$\text{Area}=\dfrac{2}{3}+\dfrac{4}{3}=2.$$

Hence, the correct answer is Option B.