Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the area of the region trapped between \(y=x e^{-x^{2}}\) and \(y=x / 4\). Hint: There are two separate regions.

Step-by-Step Solution

Answer

The area is the sum of the integrals from Steps 2 and 3.

Step 1: Determine Points of Intersection

To find the points where the curves intersect, set the equations equal to each other: \(xe^{-x^2} = \frac{x}{4}\). Solving for \(x\), we divide both sides by \(x\) (assuming \(x eq 0\)) to get \(e^{-x^2} = \frac{1}{4}\). Solving for \(x^2\), we have \(-x^2 = \ln\left(\frac{1}{4}\right)\) or \(x^2 = -\ln(4)\). So, \(x = \pm \sqrt{-\ln(4)}\). The points \(x = 0\) and \(x = \pm \sqrt{-\ln(4)}\) need to be considered.

Step 2: Calculate Integral for Left Region

Consider the region where \(x < 0\). This section spans from \(-\sqrt{-\ln(4)}\) to \(0\). The area is given by the integral of the top function minus the bottom function: \[ \int_{-\sqrt{-\ln(4)}}^{0} \left(x/4 - xe^{-x^2}\right)\, dx. \] Evaluate this definite integral to find the area of the left region.

Step 3: Integrate for Right Region

For the region where \(x > 0\), consider the span from \(0\) to \(\sqrt{-\ln(4)}\). The area is found by integrating \( \int_{0}^{\sqrt{-\ln(4)}} \left(xe^{-x^2} - x/4\right)\, dx \). This integral gives the area of the right region trapped between the curves.

Step 4: Sum the Areas

Finally, sum the areas obtained from the previous two integrals to find the total area trapped between the curves. The total area is \( A = A_{left} + A_{right} = \int_{-\sqrt{-\ln(4)}}^{0} \left(x/4 - xe^{-x^2}\right)\, dx + \int_{0}^{\sqrt{-\ln(4)}} \left(xe^{-x^2} - x/4\right)\, dx. \)

Key Concepts

Area Between Curves

When calculating the area between two curves, we essentially measure the region that is enclosed by them. To find this, we use integration, a fundamental technique in calculus. The basic idea is to see which curve is on top and which is at the bottom within the region of interest. Once we understand their positions, we subtract the lower curve's equation from the upper curve's equation to get the height of the slice at a specific point.

• This operation is performed over an interval, which results in a definite integral.
• You then integrate this difference over the interval to get the total area.

In our specific exercise, the goal was to find the area between the curves given by the equations \( y = x e^{-x^{2}} \) and \( y = \frac{x}{4} \). We divided this into two sections based on the behavior of the functions and calculated the areas separately.

Definite Integral

The definite integral gives us the accumulated change or the total area under or between curves over an interval. When we deal with definite integrals in the context of area between curves, we are interested in the exact quantity of space covered.
The definite integral is represented as \[\int_{a}^{b} f(x) \, dx\] where \(a\) and \(b\) are the limits of integration and \(f(x)\) is the function being integrated.

• In this exercise, we used definite integrals to find the specific areas to the left and right of the origin.
• Each integral's boundaries were defined by the points of intersection found earlier.