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

Short answer

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

Step-by-step solution

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.

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.

Evaluate the Area

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

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.