To find the area of the shaded region between the curves $y = 2x^2$ and $y = 16 - 4x$, we need to perform the following steps:

1. Find the points of intersection of the two curves by setting $y = 2x^2$ equal to $y = 16 - 4x$.
2. Set up the integral for the area between the curves from the left intersection point to the right intersection point.
3. Calculate the integral to find the area.

Step 1: Find the Points of Intersection

Set the equations equal to each other to find the points of intersection:

$2x^2 = 16 - 4x$

Rearrange the equation:

$2x^2 + 4x - 16 = 0$

Divide the entire equation by 2:

$x^2 + 2x - 8 = 0$

Factor the quadratic equation:

$(x + 4)(x - 2) = 0$

So, the points of intersection are $x = -4$ and $x = 2$. However, since the region shown in the graph only spans from $x = 0$ to $x = 4$, we ignore $x = -4$ and consider the interval from $x = 0$ to $x = 2$.

Step 2: Set Up the Integral

The area $A$ between the two curves is given by:

$A = \int_{a}^{b} \left[ f(x) - g(x) \right] \, dx$

Where:

• $f(x) = 16 - 4x$ (the upper function)
• $g(x) = 2x^2$ (the lower function)

The limits of integration are from $x = 0$ to $x = 2$.

$A = \int_{0}^{2} \left[(16 - 4x) - 2x^2\right] \, dx$

Step 3: Calculate the Integral

First, simplify the integrand:

$A = \int_{0}^{2} \left(16 - 4x - 2x^2\right) \, dx$

Now, integrate term by term:

$A = \left[ 16x - 2x^2 - \frac{2x^3}{3} \right]_{0}^{2}$

Evaluate at the bounds $x = 2$ and $x = 0$:

[ A = \left[ 16(2) - 2(2)^2 - \frac{2(2)^3}{3} \right] - \left[ 16(0) - 2(0)^2 - \frac{2(0)^3}{3} \right] ]

Simplify:

[ A = \left[ 32 - 8 - \frac{16}{3} \right] - [0] ]

$A = 24 - \frac{16}{3}$

To simplify further:

$A = \frac{72}{3} - \frac{16}{3} = \frac{56}{3}$

Thus, the area of the shaded region is:

$\boxed{\frac{56}{3}} \text{ square units}$

Would you like more details or have any questions?

Related Questions:

1. What is the importance of identifying the correct limits of integration in finding the area between curves?
2. How would the area change if the curves were given different functions, such as $y = x^2$ and $y = 12 - 2x$?
3. How can the method of finding the area between curves be applied to more complex shapes, such as those involving trigonometric functions?
4. What happens if the curves intersect at more than two points? How does this affect the calculation of the area?
5. How do you determine whether one curve is above another over a specific interval?

Tip:

When solving integrals involving areas between curves, always graph the functions first if possible. This visualization helps in identifying the correct limits of integration and ensures that you subtract the lower curve from the upper curve.