The problem asks to find the area enclosed between the two functions $f(x) = 0.3x^2 + 5$ and $g(x) = x$, over the interval $x = -2$ to $x = 7$.

To solve this, the area between two curves $f(x)$ and $g(x)$ over an interval $[a, b]$ is given by:

$\text{Area} = \int_{a}^{b} |f(x) - g(x)| \, dx$

Step-by-step solution:

1. Find the difference between the two functions: $f(x) - g(x) = (0.3x^2 + 5) - x = 0.3x^2 - x + 5$

2. Set up the integral for the area: Since we are calculating the area from $x = -2$ to $x = 7$, we set up the integral:

   $\text{Area} = \int_{-2}^{7} \left( 0.3x^2 - x + 5 \right) dx$

3. Evaluate the integral:

   • The integral of $0.3x^2$ is $\frac{0.3x^3}{3} = 0.1x^3$
   • The integral of $-x$ is $-\frac{x^2}{2}$
   • The integral of $5$ is $5x$

   So, the indefinite integral is:

   $\int \left( 0.3x^2 - x + 5 \right) dx = 0.1x^3 - \frac{x^2}{2} + 5x$

4. Evaluate the definite integral:

   Now, substitute the limits $x = 7$ and $x = -2$ into the integral:

   [ \text{Area} = \left[ 0.1(7^3) - \frac{7^2}{2} + 5(7) \right] - \left[ 0.1(-2^3) - \frac{(-2)^2}{2} + 5(-2) \right] ]

   Let me calculate that for you.The area enclosed between the curves $f(x) = 0.3x^2 + 5$ and $g(x) = x$ from $x = -2$ to $x = 7$ is $57.6$ square units.

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:

1. How do we calculate the area between two curves when they intersect?
2. What changes if the interval of integration is different?
3. How can we determine the points of intersection of $f(x)$ and $g(x)$?
4. How does the method change if the curves cross within the interval?
5. Can this method be applied to functions with vertical asymptotes?

Tip: Always check if the curves intersect within the interval, as it may require splitting the integral at the points of intersection.