To solve the problem of finding the area bounded by the given equations $y = e^x$ and $y = -\frac{1}{x}$ over the interval $1 \leq x \leq 2$, we proceed as follows:

1. Find the points of intersection (if necessary): For $y = e^x$ and $y = -\frac{1}{x}$, we check if they intersect in the given interval $1 \leq x \leq 2$. However, in this case, we are simply calculating the bounded area between the two curves over the interval, so intersection points are not required here.

2. Set up the integral:
   The upper function is $y = e^x$, and the lower function is $y = -\frac{1}{x}$. The area $A$ is given by: $A = \int_1^2 \left(e^x - \left(-\frac{1}{x}\right)\right) dx = \int_1^2 \left(e^x + \frac{1}{x}\right) dx.$

3. Evaluate the integral: Break it into two parts: $A = \int_1^2 e^x dx + \int_1^2 \frac{1}{x} dx.$

   • The first integral: $\int e^x dx = e^x \quad \text{evaluated from 1 to 2.}$ This becomes: $e^2 - e^1.$

   • The second integral: $\int \frac{1}{x} dx = \ln|x| \quad \text{evaluated from 1 to 2.}$ This becomes: $\ln(2) - \ln(1).$ Since $\ln(1) = 0$, this simplifies to $\ln(2)$.

4. Combine the results: Add the results of both integrals: $A = (e^2 - e^1) + \ln(2).$ Simplify: $A = e^2 - e + \ln(2).$

Thus, the bounded area is $A = e^2 - e + \ln(2)$.

Let me know if you would like me to provide further clarification, a numerical approximation of the result, or visualize the bounded region.

Here are 5 related questions you may find helpful:

1. How do you graphically verify the bounded area between two curves?
2. What is the difference between definite and indefinite integration?
3. How does the choice of interval affect the bounded area calculation?
4. How is the natural exponential function $e^x$ important in calculus?
5. How does logarithmic integration apply to real-world problems?

Tip: Always ensure the correct order of subtraction (upper function minus lower function) when setting up the integral for bounded areas.