let the region r be the area enclosed by the function f(x)=3e^x and g(x)=4x + 3. if the region r is the base…

let the region r be the area enclosed by the function f(x)=3e^x and g(x)=4x + 3. if the region r is the base of a solid such that each cross - section perpendicular to the x - axis is a rectangle whose height is twice the length of its base in the region r, find the volume of the solid. you may use a calculator and round to the nearest thousandth.

let the region r be the area enclosed by the function f(x)=3e^x and g(x)=4x + 3. if the region r is the base of a solid such that each cross - section perpendicular to the x - axis is a rectangle whose height is twice the length of its base in the region r, find the volume of the solid. you may use a calculator and round to the nearest thousandth.

Answer

Explanation:

Step1: Find intersection points

Set $3e^{x}=4x + 3$. Using a calculator, the intersection points are approximately $x_1 = 0$ and $x_2\approx0.91$.

Step2: Determine base of rectangle

The base of each rectangle perpendicular to the $x -$axis is $b=f(x)-g(x)=3e^{x}-(4x + 3)$.

Step3: Determine height of rectangle

The height $h$ of the rectangle is $h = 2b=2(3e^{x}-(4x + 3))$.

Step4: Set up volume integral

The volume $V$ of the solid using the cross - sectional area formula $V=\int_{a}^{b}A(x)dx$, where $A(x)$ is the area of the cross - section. Here, $A(x)=b\times h=(3e^{x}-(4x + 3))\times2(3e^{x}-(4x + 3))=2(3e^{x}-4x - 3)^{2}$. So $V = \int_{0}^{0.91}2(3e^{x}-4x - 3)^{2}dx$.

Step5: Evaluate integral

Using a calculator to evaluate $\int_{0}^{0.91}2(3e^{x}-4x - 3)^{2}dx$, we get $V\approx2.457$.

Answer:

$2.457$