To find the area between the curves \( f(x) = \frac{1}{2}x^2 + 1 \) and \( g(x) = -x^2 + \frac{3}{2}x + 4 \), we first need to determine the points where these curves intersect. Set \( f(x) \) equal to \( g(x) \):
\[
\frac{1}{2}x^2 + 1 = -x^2 + \frac{3}{2}x + 4
\]
Combine like terms and solve the quadratic equation:
\[
\frac{1}{2}x^2 + x^2 - \frac{3}{2}x + 1 - 4 = 0
\]
\[
\frac{3}{2}x^2 - \frac{3}{2}x - 3 = 0
\]
Multiply through by 2 to clear the fraction:
\[
3x^2 - 3x - 6 = 0
\]
Divide by 3:
\[
x^2 - x - 2 = 0
\]
Factor the quadratic equation:
\[
(x - 2)(x + 1) = 0
\]
So, the intersection points are \( x = 2 \) and \( x = -1 \).
The area \( A \) between the curves from \( x = -1 \) to \( x = 2 \) is given by the integral of the difference between \( g(x) \) and \( f(x) \):
\[
A = \int_{-1}^{2} [g(x) - f(x)] \, dx
\]
Substitute the functions:
\[
A = \int_{-1}^{2} \left[ \left( -x^2 + \frac{3}{2}x + 4 \right) - \left( \frac{1}{2}x^2 + 1 \right) \right] \, dx
\]
Simplify the integrand:
\[
A = \int_{-1}^{2} \left( -x^2 + \frac{3}{2}x + 4 - \frac{1}{2}x^2 - 1 \right) \, dx
\]
\[
A = \int_{-1}^{2} \left( -\frac{3}{2}x^2 + \frac{3}{2}x + 3 \right) \, dx
\]
Integrate term by term:
\[
A = \int_{-1}^{2} -\frac{3}{2}x^2 \, dx + \int_{-1}^{2} \frac{3}{2}x \, dx + \int_{-1}^{2} 3 \, dx
\]
\[
A = -\frac{3}{2} \int_{-1}^{2} x^2 \, dx + \frac{3}{2} \int_{-1}^{2} x \, dx + 3 \int_{-1}^{2} 3 \, dx
\]
Calculate each integral:
\[
-\frac{3}{2} \left[ \frac{x^3}{3} \right]_{-1}^{2} + \frac{3}{2} \left[ \frac{x^2}{2} \right]_{-1}^{2} + 3 \left[ x \right]_{-1}^{2}
\]
\[
-\frac{3}{2} \left( \frac{2^3}{3} - \frac{(-1)^3}{3} \right) + \frac{3}{2} \left( \frac{2^2}{2} - \frac{(-1)^2}{2} \right) + 3 \left( 2 - (-1) \right)
\]
\[
-\frac{3}{2} \left( \frac{8}{3} + \frac{1}{3} \right) + \frac{3}{2} \left( 2 - \frac{1}{2} \right) + 3 \times 3
\]
\[
-\frac{3}{2} \times 3 + \frac{3}{2} \times \frac{3}{2} + 9
\]
\[
-4.5 + 2.25 + 9
\]
\[
6.75
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