To find the area bounded by the graphs of the two equations $y = e^{0.75x}$ and $y = -\frac{1}{x}$ over the interval $1 \leq x \leq 2$, we need to follow these steps:

Step 1: Set up the integral

The area between two curves is given by the formula: $A = \int_{a}^{b} \left( f(x) - g(x) \right) \, dx$ where $f(x)$ is the upper function and $g(x)$ is the lower function over the interval $[a, b]$.

From the problem:

• The function $f(x) = e^{0.75x}$
• The function $g(x) = -\frac{1}{x}$

We are given the interval $1 \leq x \leq 2$.

Step 2: Identify the upper and lower functions

To find out which function is on top, we need to evaluate both functions at a few points within the interval:

• At $x = 1$: $e^{0.75(1)} = e^{0.75} \approx 2.117$ $-\frac{1}{1} = -1$ So, $e^{0.75x}$ is greater than $-\frac{1}{x}$ at $x = 1$.

• At $x = 2$: $e^{0.75(2)} = e^{1.5} \approx 4.4817$ $-\frac{1}{2} = -0.5$ So, $e^{0.75x}$ is also greater than $-\frac{1}{x}$ at $x = 2$.

Thus, the function $f(x) = e^{0.75x}$ is the upper function, and $g(x) = -\frac{1}{x}$ is the lower function.

Step 3: Set up the integral

Now we set up the integral for the area between the curves: $A = \int_{1}^{2} \left( e^{0.75x} - \left( -\frac{1}{x} \right) \right) \, dx$ which simplifies to: $A = \int_{1}^{2} \left( e^{0.75x} + \frac{1}{x} \right) \, dx$

Step 4: Compute the integral

We need to compute the integral of both terms separately:

1. The integral of $e^{0.75x}$ is: $\int e^{0.75x} \, dx = \frac{1}{0.75} e^{0.75x} = \frac{4}{3} e^{0.75x}$
2. The integral of $\frac{1}{x}$ is: $\int \frac{1}{x} \, dx = \ln|x|$

Now, we apply the limits of integration from 1 to 2.

$A = \left[ \frac{4}{3} e^{0.75x} + \ln|x| \right]_{1}^{2}$ Substitute the upper limit $x = 2$ and the lower limit $x = 1$:

$A = \left( \frac{4}{3} e^{0.75(2)} + \ln(2) \right) - \left( \frac{4}{3} e^{0.75(1)} + \ln(1) \right)$ Since $\ln(1) = 0$, this simplifies to: $A = \left( \frac{4}{3} e^{1.5} + \ln(2) \right) - \frac{4}{3} e^{0.75}$

Now let's compute this numerically:

$A \approx \left( \frac{4}{3} \times 4.4817 + 0.6931 \right) - \frac{4}{3} \times 2.117$

$A \approx \left( 5.9756 + 0.6931 \right) - 2.8227$

$A \approx 6.6687 - 2.8227 = 3.846$

Final Answer:

The area is approximately $3.846$ square units.

Let me know if you want further details or have any questions!

Here are 5 questions related to this:

1. How would you interpret the area between curves geometrically?
2. What would change if the interval was $1 \leq x \leq 3$?
3. How does the exponential growth of $e^{0.75x}$ affect the area?
4. Can we find the exact intersection point of the two curves?
5. How do we handle areas when curves cross over each other?

Tip: When calculating the area between curves, always make sure you correctly identify which curve is on top!