To find the area of the region bounded by the curves [tex]\( y = 5x^2 \)[/tex] and [tex]\( y = 3x^2 + 8 \)[/tex], we can follow these steps:
1. Identify the Points of Intersection:
To find the points where the curves intersect, set the equations equal to each other:
[tex]\[
5x^2 = 3x^2 + 8
\][/tex]
Rearrange this to:
[tex]\[
5x^2 - 3x^2 = 8
\][/tex]
Simplify:
[tex]\[
2x^2 = 8
\][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[
x^2 = 4 \implies x = \pm 2
\][/tex]
Thus, the points of intersection are [tex]\( x = -2 \)[/tex] and [tex]\( x = 2 \)[/tex].
2. Set Up the Integral:
The area between two curves from [tex]\( x = a \)[/tex] to [tex]\( x = b \)[/tex] is given by the integral of the difference of the functions:
[tex]\[
\text{Area} = \int_{a}^{b} [f(x) - g(x)] \, dx
\][/tex]
Here, [tex]\( f(x) = 3x^2 + 8 \)[/tex] and [tex]\( g(x) = 5x^2 \)[/tex]. The integral becomes:
[tex]\[
\int_{-2}^{2} [(3x^2 + 8) - 5x^2] \, dx
\][/tex]
3. Simplify the Integrand:
Simplify inside the integral:
[tex]\[
\int_{-2}^{2} (3x^2 + 8 - 5x^2) \, dx = \int_{-2}^{2} (-2x^2 + 8) \, dx
\][/tex]
4. Evaluate the Integral:
Split the integral into two parts:
[tex]\[
\int_{-2}^{2} (-2x^2 + 8) \, dx = \int_{-2}^{2} -2x^2 \, dx + \int_{-2}^{2} 8 \, dx
\][/tex]
Evaluate each part separately:
[tex]\[
\int_{-2}^{2} -2x^2 \, dx = -2 \int_{-2}^{2} x^2 \, dx \quad \text{(We know that the integral of an even function over symmetric limits is twice the integral from 0 to upper limit)}
\][/tex]
[tex]\[
= -2 \left[ \frac{x^3}{3} \right]_{-2}^{2} = -2 \left( \frac{2^3}{3} - \frac{(-2)^3}{3} \right)
\][/tex]
[tex]\[
= -2 \left( \frac{8}{3} - \left( \frac{-8}{3} \right) \right) = -2 \left( \frac{8 + 8}{3} \right) = -2 \left( \frac{16}{3} \right) = -\frac{32}{3}
\][/tex]
Next, for the constant term:
[tex]\[
\int_{-2}^{2} 8 \, dx = 8 \int_{-2}^{2} \, dx = 8 [x]_{-2}^{2} = 8 \left( 2 - (-2) \right) = 8 \cdot 4 = 32
\][/tex]
Combine the results:
[tex]\[
\text{Area} = -\frac{32}{3} + 32 = -\frac{32}{3} + \frac{96}{3} = \frac{-32 + 96}{3} = \frac{64}{3}
\][/tex]
So, the area of the region bounded by the curves [tex]\( y = 5x^2 \)[/tex] and [tex]\( y = 3x^2 + 8 \)[/tex] is:
[tex]\[
\boxed{\frac{64}{3}}
\][/tex]
