Problem 39 - College Algebra and Calculus: An Applied Approach

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. $$ y=\frac{x}{1+e^{x^{2}}}, y=0, x=2 $$

Step-by-Step Solution

Answer

The exact area under the curve is 2. The graphing utility also shows an area close to 2, confirming our calculations.

Step 1: Evaluate the Definite Integral

The area A between the x-axis (y=0) and the curve \(y=\frac{x}{1+e^{x^{2}}}\) from \(x=0\) to \(x=2\) is given by: \[A = \int_0^2 \frac{x}{1+e^{x^{2}}} dx\]To evaluate this integral, it might be useful to make a variable substitution due to the presence of the exponential term in the denominator. Let \(u = x^{2}\), then \(du = 2x dx\), and \(x dx = \frac{1}{2} du\). The limits of integration will also change accordingly: for \(x=0\) we have \(u=0\), and for \(x=2\) we have \(u=4\). Therefore, the integral becomes: \[A = \frac{1}{2} \int_0^4 \frac{1}{1+e^{u}} du\]This is a simple integral to solve using direct integration techniques.

Step 2: Solve the Integral

The integral in step one is a standard integral of the form \( \int \frac{1}{a+e^x} dx = ln|a+e^x| + C \). Applying this property:\[A = \frac{1}{2} ln|1+e^{u}| \Bigg|_0^4\]Simplifying this expression by substituting the limits back in should give us the exact area under the curve.

Step 3: Evaluate the Area

Substituting the limits of integration into our result from step 2, we find:\[A = \frac{1}{2} (ln(1+e^{4}) - ln(1+e^{0}))= \frac{1}{2} ln(e^{4}) = 2 \]

Step 4: Approximate Area Using Graphing Utility

To approximate the area under the curve, one can use a graphing utility. Graph the function \(y=\frac{x}{1+e^{x^{2}}}\) from \(x=0\) to \(x=2\) and use the tool's 'Area under Curve' function to get the area. The result should be very close to 2, which confirms the calculation in steps 2 and 3.