let the region r be the area enclosed the function f(x)=e^x, the horizontal line y = 17, and the y - axis…

Explanation:

Step1: Find the intersection - point of $y = e^{x}$ and $y = 17$

Set $e^{x}=17$, then $x=\ln(17)\approx2.833$.

Step2: Integral in terms of $x$

The region is bounded by $x = 0$ and $x=\ln(17)$ with the upper - curve $y = 17$ and the lower - curve $y = e^{x}$. The area formula for a region bounded by two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is $A=\int_{a}^{b}(f(x)-g(x))dx$. Here, $a = 0$, $b=\ln(17)$, $f(x)=17$, and $g(x)=e^{x}$. So the integral in terms of $x$ is $\int_{0}^{\ln(17)}(17 - e^{x})dx$.

Step3: Express $x$ in terms of $y$

Since $y = e^{x}$, then $x=\ln(y)$. The region is bounded by $y = 1$ (when $x = 0$, $y=e^{0}=1$) and $y = 17$ with the right - curve $x=\ln(y)$ and the left - curve $x = 0$. The area formula for a region bounded by two curves $x = h(y)$ and $x = k(y)$ from $y=c$ to $y = d$ is $A=\int_{c}^{d}(h(y)-k(y))dy$. Here, $c = 1$, $d = 17$, $h(y)=\ln(y)$, and $k(y)=0$. So the integral in terms of $y$ is $\int_{1}^{17}\ln(y)dy$.

Answer:

$x_1 = 0$ $x_2=\ln(17)\approx2.833$ $\int_{0}^{\ln(17)}(17 - e^{x})dx$ $y_1 = 1$ $y_2 = 17$ $\int_{1}^{17}\ln(y)dy$