The problem asks us to find the volume of a pyramid with a height of 4 meters, given that the cross-sectional area at a height $x$ is $A(x) = 2x^2$. We are given the formula for the volume of a solid as an integral of the cross-sectional area, $Volume = \int_a^b A(x) dx$.

1. Problem Description

The problem asks us to find the volume of a pyramid with a height of 4 meters, given that the cross-sectional area at a height $x$ is $A(x) = 2x^2$. We are given the formula for the volume of a solid as an integral of the cross-sectional area, $Volume = \int_a^b A(x) dx$.

2. Solution Steps

To find the volume of the pyramid, we need to integrate the cross-sectional area function $A(x) = 2x^2$ from $x=0$ to $x=4$, where 4 is the height of the pyramid.

$V = \int_0^4 A(x) dx = \int_0^4 2x^2 dx$

First, we find the antiderivative of $2x^2$:

$\int 2x^2 dx = \frac{2x^3}{3} + C$

Next, we evaluate the definite integral:

$V = \left[ \frac{2x^3}{3} \right]_0^4 = \frac{2(4^3)}{3} - \frac{2(0^3)}{3} = \frac{2(64)}{3} - 0 = \frac{128}{3}$

3. Final Answer

The volume of the pyramid is $\frac{128}{3}$ cubic meters.