Question

Use the table of integrals to find the exact area of the region bounded by the graphs of the equations. Then use a graphing utility to graph the region and approximate the area. $$ y=\frac{-e^{x}}{1-e^{2 x}}, y=0, x=1, x=2 $$

Short answer

The exact area can be found by integrating the function \( y = \frac{-e^{x}}{1-e^{2 x}} \) over the interval \([1, 2]\). Due to the complexity of the integral, the exact answer is not readily available. However, the exact answer can be obtained by following the detailed steps, using the right methods and then checking the results with a suitable graphing utility.

Step-by-step solution

Define the Integral

Set up the integral which will give the area under the curve. The interval of integration is from \( x = 1 \) to \( x = 2 \). So, the integral is \(\int_{1}^{2}(-e^{x})/(1 - e^{2x}) dx\).

Find the Antiderivative

Use an applicable rule from the table of integrals to find the antiderivative of the function. In this case, it may be challenging to find a direct match, but the expression in the integral can be rewritten as \(-e^{x} * (e^{-2x})/(1 - e^{2x} * e^{-2x})\) which simplifies to \(-e^{-x}/(e^{-x} - e^{-x + 2})\). The antiderivative can then be found. The antiderivative should be evaluated at x = 2 and x = 1 to find the exact area.

Evaluate the Antiderivative at the limits

Evaluate the antiderivative at the upper limit and lower limit of the integral, that is, at x = 2 and x = 1 respectively. The difference will give the exact area under the curve from x = 1 to x = 2.

Use a graphing utility to check the answer

Plot the function \( y = \frac{-e^{x}}{1-e^{2 x}} \) from x = 1 to x = 2 on a graphing utility and use the utility's integral feature to verify the result.