To find the net area between the graph of [tex]\( f(x) = x^3 + 3x^2 - 2x + 3 \)[/tex] and the [tex]\( x \)[/tex]-axis over the interval [tex]\([-2, 2]\)[/tex], we need to evaluate the definite integral of [tex]\( f(x) \)[/tex] from [tex]\(-2\)[/tex] to [tex]\(2\)[/tex].
The definite integral:
[tex]\[ \int_{-2}^{2} (x^3 + 3x^2 - 2x + 3) \, dx \][/tex]
represents the net area.
Let's proceed step-by-step:
1. Set up the integral:
[tex]\[ \int_{-2}^{2} (x^3 + 3x^2 - 2x + 3) \, dx \][/tex]
2. Evaluate the integral:
To integrate each term independently, we find the antiderivative of [tex]\( f(x) \)[/tex]:
[tex]\[
\int x^3 \, dx = \frac{x^4}{4}, \quad \int 3x^2 \, dx = x^3, \quad \int (-2x) \, dx = -x^2, \quad \int 3 \, dx = 3x
\][/tex]
Thus, the antiderivative of [tex]\( f(x) \)[/tex] is:
[tex]\[
F(x) = \frac{x^4}{4} + x^3 - x^2 + 3x
\][/tex]
3. Evaluate [tex]\( F(x) \)[/tex] at the bounds:
[tex]\[ F(2) = \frac{2^4}{4} + 2^3 - 2^2 + 3 \cdot 2 \][/tex]
[tex]\[ F(2) = \frac{16}{4} + 8 - 4 + 6 \][/tex]
[tex]\[ F(2) = 4 + 8 - 4 + 6 \][/tex]
[tex]\[ F(2) = 14 \][/tex]
[tex]\[ F(-2) = \frac{(-2)^4}{4} + (-2)^3 - (-2)^2 + 3 \cdot (-2) \][/tex]
[tex]\[ F(-2) = \frac{16}{4} - 8 - 4 - 6 \][/tex]
[tex]\[ F(-2) = 4 - 8 - 4 - 6 \][/tex]
[tex]\[ F(-2) = -14 \][/tex]
4. Subtract the values:
[tex]\[ \int_{-2}^{2} (x^3 + 3x^2 - 2x + 3) \, dx = F(2) - F(-2) \][/tex]
[tex]\[ = 14 - (-14) \][/tex]
[tex]\[ = 14 + 14 \][/tex]
[tex]\[ = 28 \][/tex]
So the net area between the graph of [tex]\( f(x) \)[/tex] and the [tex]\( x \)[/tex]-axis over the interval [tex]\([-2, 2]\)[/tex] is [tex]\( 28.0 \)[/tex] (rounded to two decimal places).
Answer:
28.0
