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…

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 or software, the intersection points are approximately $x_1 = 0$ and $x_2\approx1.146$.

Step2: Determine base of cross - section

The base of each cross - section perpendicular to the $x$ - axis is $b=\vert3e^{x}-(4x + 3)\vert$. Since $3e^{x}\geq4x + 3$ on the interval of interest, $b = 3e^{x}-(4x + 3)$.

Step3: Determine height of cross - section

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

Step4: Find area of cross - section

The area of the cross - section $A(x)=b\times h=(3e^{x}-(4x + 3))\times2(3e^{x}-(4x + 3))=2(3e^{x}-4x - 3)^{2}$.

Step5: Calculate volume using integral

The volume $V=\int_{0}^{1.146}A(x)dx=\int_{0}^{1.146}2(3e^{x}-4x - 3)^{2}dx$. Expand $(3e^{x}-4x - 3)^{2}=9e^{2x}-24xe^{x}-18e^{x}+16x^{2}+24x + 9$. Then $V = 2\int_{0}^{1.146}(9e^{2x}-24xe^{x}-18e^{x}+16x^{2}+24x + 9)dx$. We know that:

  • $\int e^{2x}dx=\frac{1}{2}e^{2x}+C$, $\int xe^{x}dx=(x - 1)e^{x}+C$, $\int e^{x}dx=e^{x}+C$, $\int x^{2}dx=\frac{1}{3}x^{3}+C$, $\int xdx=\frac{1}{2}x^{2}+C$, $\int 1dx=x+C$. $V = 2\left[9\times\frac{1}{2}e^{2x}-24((x - 1)e^{x})-18e^{x}+16\times\frac{1}{3}x^{3}+24\times\frac{1}{2}x^{2}+9x\right]_{0}^{1.146}$. Evaluating the definite integral: [ \begin{align*} V&=2\left[\left(\frac{9}{2}e^{2\times1.146}-24((1.146 - 1)e^{1.146})-18e^{1.146}+\frac{16}{3}(1.146)^{3}+12(1.146)^{2}+9(1.146)\right)-\left(\frac{9}{2}-0 - 18+0 + 0+0\right)\right]\ &\approx2\left[\left(\frac{9}{2}e^{2.292}-24\times0.146e^{1.146}-18e^{1.146}+\frac{16}{3}(1.146)^{3}+12(1.146)^{2}+9(1.146)\right)-\left(\frac{9}{2}-18\right)\right]\ &\approx2\left[\left(\frac{9}{2}\times9.917-24\times0.146\times3.146-18\times3.146+\frac{16}{3}\times1.509+12\times1.313+9\times1.146\right)-\left(4.5 - 18\right)\right]\ &\approx2\left[\left(44.6265-10.877 - 56.628+8.05+15.756+10.314\right)+13.5\right]\ &\approx2\left[(44.6265 + 8.05+15.756+10.314-(10.877 + 56.628))+13.5\right]\ &\approx2\left[(78.7525 - 67.505)+13.5\right]\ &\approx2\left[11.2475+13.5\right]\ &\approx2\times24.7475\ &\approx49.495 \end{align*} ]

Answer:

$49.495$