question let the region r be the area enclosed by the function f(x)=e^x, the horizontal line y = 7, and the…

Explanation:

Step1: Find the intersection point

Set $e^{x}=7$, then $x = \ln(7)$.

Step2: Determine the base of the isosceles - right - triangle cross - section

The length of the leg of the isosceles right - triangle cross - section perpendicular to the $x$ - axis is $b=7 - e^{x}$.

Step3: Find the area formula of the cross - section

The area formula of an isosceles right - triangle is $A=\frac{1}{2}s^{2}$, where $s$ is the length of the leg. So, $A(x)=\frac{1}{2}(7 - e^{x})^{2}=\frac{1}{2}(49-14e^{x}+e^{2x})$.

Step4: Calculate the volume using the integral

The volume $V$ of the solid with cross - sectional area $A(x)$ from $x = 0$ to $x=\ln(7)$ is given by the integral $V=\int_{0}^{\ln(7)}A(x)dx=\int_{0}^{\ln(7)}\frac{1}{2}(49 - 14e^{x}+e^{2x})dx$. We know that $\int kdx=kx + C$ ($k$ is a constant), $\int e^{x}dx=e^{x}+C$ and $\int e^{ax}dx=\frac{1}{a}e^{ax}+C$ ($a\neq0$). [ \begin{align*} V&=\frac{1}{2}\left[49x-14e^{x}+\frac{1}{2}e^{2x}\right]_{0}^{\ln(7)}\ &=\frac{1}{2}\left[\left(49\ln(7)-14\times7+\frac{1}{2}\times49\right)-\left(0 - 14+\frac{1}{2}\right)\right]\ &=\frac{1}{2}\left[49\ln(7)-98+\frac{49}{2}+14-\frac{1}{2}\right]\ &=\frac{1}{2}\left[49\ln(7)-98 + 14+\frac{49 - 1}{2}\right]\ &=\frac{1}{2}\left[49\ln(7)-84 + 24\right]\ &=\frac{1}{2}\left[49\ln(7)-60\right]\ &\approx\frac{1}{2}(49\times1.946 - 60)\ &\approx\frac{1}{2}(95.354-60)\ &\approx17.677 \end{align*} ]

Answer:

$17.677$