Find the area $S$ enclosed by the two parabolas $y = 2x^2 + 1$ and $y = x^2 - 2x + 4$.

1. Problem Description

Find the area $S$ enclosed by the two parabolas $y = 2x^2 + 1$ and $y = x^2 - 2x + 4$.

2. Solution Steps

First, we need to find the points of intersection of the two parabolas by setting the equations equal to each other:

$2x^2 + 1 = x^2 - 2x + 4$

$2x^2 - x^2 + 2x + 1 - 4 = 0$

$x^2 + 2x - 3 = 0$

$(x + 3)(x - 1) = 0$

$x = -3$ or $x = 1$.

So, the intersection points are $x = -3$ and $x = 1$.

Now we need to find the area between the curves. The area is given by the integral of the absolute difference between the two functions over the interval $[-3, 1]$.

$S = \int_{-3}^{1} |(2x^2 + 1) - (x^2 - 2x + 4)| dx$

$S = \int_{-3}^{1} |2x^2 + 1 - x^2 + 2x - 4| dx$

$S = \int_{-3}^{1} |x^2 + 2x - 3| dx$

Since $x^2 + 2x - 3 = (x+3)(x-1)$, the expression is negative between $-3$ and $1$. So, we change the sign inside the integral.

$S = \int_{-3}^{1} -(x^2 + 2x - 3) dx$

$S = \int_{-3}^{1} (-x^2 - 2x + 3) dx$

$S = [-\frac{1}{3}x^3 - x^2 + 3x]_{-3}^{1}$

$S = (-\frac{1}{3}(1)^3 - (1)^2 + 3(1)) - (-\frac{1}{3}(-3)^3 - (-3)^2 + 3(-3))$

$S = (-\frac{1}{3} - 1 + 3) - (-\frac{1}{3}(-27) - 9 - 9)$

$S = (-\frac{1}{3} + 2) - (9 - 9 - 9)$

$S = \frac{-1 + 6}{3} - (-9)$

$S = \frac{5}{3} + 9$

$S = \frac{5}{3} + \frac{27}{3}$

$S = \frac{32}{3}$

3. Final Answer

The area enclosed by the two parabolas is $\frac{32}{3}$.