Evaluate ∬_D (x + 2y) dA, where D is the region bounded by the parabolas y = 2x² and y = 1 + x².

Steps to Solve

1. Find Points of Intersection

To determine the area ( D ) bounded by the parabolas, we need to find the points where the equations intersect:

Set ( 2x^2 = 1 + x^2 ).

This simplifies to: $$ 2x^2 - x^2 - 1 = 0 \Rightarrow x^2 - 1 = 0 $$ Thus, ( x = \pm 1 ).

2. Set up the Limits for the Integral

The region ( D ) is vertically bounded by the parabolas from ( x = -1 ) to ( x = 1 ). The upper curve is ( y = 1 + x^2 ) and the lower curve is ( y = 2x^2 ).

We can set up the double integral as follows: $$ \int_{-1}^{1} \int_{2x^2}^{1+x^2} (x + 2y) , dy , dx $$

3. Evaluate the Inner Integral

Evaluate the inner integral: $$ \int_{2x^2}^{1+x^2} (x + 2y) , dy $$ The integral of ( (x + 2y) ) with respect to ( y ) is: $$ xy + y^2 $$ Calculating the bounds: $$ \left[ x(1+x^2) + (1+x^2)^2 \right] - \left[ x(2x^2) + (2x^2)^2 \right] $$

Calculating it requires simplifying:

1. Upper bound: $$ x(1+x^2) + (1+x^2)^2 = x + x^3 + 1 + 2x^2 + x^4 $$
2. Lower bound: $$ x(2x^2) + (2x^2)^2 = 2x^3 + 4x^4 $$

Thus, the difference is: $$ x + x^3 + 1 + 2x^2 + x^4 - (2x^3 + 4x^4) = x + 1 + 2x^2 - x^3 - 3x^4 $$

4. Evaluate the Outer Integral

Now, we need to integrate: $$ \int_{-1}^{1} (x + 1 + 2x^2 - x^3 - 3x^4) , dx $$ Break it into individual integrals: $$ \int_{-1}^{1} x , dx + \int_{-1}^{1} 1 , dx + 2\int_{-1}^{1} x^2 , dx - \int_{-1}^{1} x^3 , dx - 3\int_{-1}^{1} x^4 , dx $$

Using symmetry:

• The integrals of odd functions ( \int_{-1}^{1} x , dx ) and ( \int_{-1}^{1} x^3 , dx ) are zero.
• The integrals of even functions ( \int_{-1}^{1} x^2 , dx = \frac{2}{3} ) and ( \int_{-1}^{1} x^4 , dx = \frac{2}{5} ).

Thus, we have: $$ 0 + 2 + 2(\frac{2}{3}) - 0 - 3(\frac{2}{5}) = 2 + \frac{4}{3} - \frac{6}{5} $$

Finding a common denominator (15): $$ 2 = \frac{30}{15}, \quad \frac{4}{3} = \frac{20}{15}, \quad \frac{6}{5} = \frac{18}{15} $$ So, $$ \frac{30 + 20 - 18}{15} = \frac{32}{15} $$

5. Final Answer

Thus, the value of the integral is: $$ \frac{32}{15} $$