Problem 39 - Essential Calculus

Question

Sketch the region bounded by the graphs of the functions, and find the area of the region. $$ f(x)=x e^{-x^{2}}, \quad y=0, \quad 0 \leq x \leq 1 $$

Step-by-Step Solution

Answer

The area under the curve \(f(x) = x e^{-x^2}\) from \(x=0\) to \(x=1\) is \(\frac{2-e}{2e}\).

Step 1: Sketch the given function.

To gain a more comprehensive understanding of the question, sketch the given function \(f(x) = x e^{-x^2}\) for the provided interval, which is \(0 \leq x \leq 1\). In this region the function is positive and the graph lies above the x-axis.

Step 2: Identify the bounded region.

The bounded region is the area between the curve \(f(x) = x e^{-x^2}\) and the x-axis (y=0) from \(x=0\) to \(x=1\). This is the area we are asked to compute.

Step 3: Compute the area under the curve.

To find the area under a curve from \(x=a\) to \(x=b\), we can use the definite integral from \(a\) to \(b\) of the function. Apply this concept to the bounded region from Step 2, the integral to compute the area under the curve is \(\int_{0}^{1} x e^{-x^{2}} dx\). This can be found using a simple substitution \(u=x^2\), \(du=2x dx\) and changing the bounds accordingly. The resulting integral \(\int_{0}^{1} \frac{e^{-u}}{2} du\) can be integrated using standard techniques of calculating the integral of exponential functions.

Step 4: Evaluate the definite integral.

After performing the calculations from the previous steps, we get \(\left. -\frac{e^{-u}}{2} \right|_0^1 = -\frac{1}{2e} + \frac{1}{2} = \frac{2-e}{2e}\).

Key Concepts

Definite Integral

The concept of a definite integral is all about calculating the total area under a curve between two specified points on the x-axis. In this exercise, we are finding the area under the curve of the function \( f(x) = x e^{-x^2} \) between \( x = 0 \) and \( x = 1 \). The definite integral is represented by the integral sign with limits, \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper boundaries, respectively.
The process involves finding an antiderivative (which is a function whose derivative gives us back the original function) and then evaluating this at the boundary points. This gives us a single number representing the total area over that interval.

• Definite integrals are used for calculating areas, displacement, and accumulated quantities.
• They provide a geometric representation, especially useful in physics and engineering.

Area Under a Curve

Finding the area under a curve is a central application of calculus, and it is a direct use of definite integrals. In our exercise, the curve is given by \( f(x) = x e^{-x^2} \), and we’re interested in the region confined between this curve and the x-axis, from \( x = 0 \) to \( x = 1 \).
To achieve this, we calculate the definite integral of the function across the interval. This integral represents the sum of infinite small rectangles that fit perfectly under the curve. It's like slicing the area into infinite tiny slices that perfectly conform to the curve's shape.

• The area under a curve between two points gives us a precise measure of the space between the curve and the baseline (in our case, the x-axis).
• This technique can also be used to calculate probabilities in statistics and total quantities in science.

Exponential Function

Exponential functions play a crucial role in calculus, and they're often encountered in growth and decay problems. In this exercise, we see the function component \( e^{-x^2} \), which is part of the exponential family, characterized by \( e \), the base of the natural logarithm (approximately 2.718).
Exponential functions are unique due to their constant rate of growth or decay. They increase or decrease by a consistent factor rather than a set amount. This property makes them suitable for modeling many natural processes, such as population growth or radioactive decay.

• A common distinctive feature is the inclusion of \( e \) in the formulation. Functions of this form are ubiquitous in calculus.
• The decay in \( e^{-x^2} \) results from the negative exponent, affecting how the function behaves graphically, rapidly decreasing as \( x \) grows.

Integration Techniques

Various techniques can be employed to solve integrals, which are essential in finding areas under curves and solving differential equations. In this exercise, a substitution technique simplifies the integral of \( x e^{-x^2} \).
The substitution method involves changing variables to transform a complex integral into a simpler one. Here, we substitute \( u = x^2 \), making \( du = 2x \, dx \), which simplifies our original integral. This transformation aids in converting a difficult integration process into something more manageable.

• Choosing the right substitution can greatly simplify the integration process and is key in solving intricate problems effectively.
• Other integration techniques may include integration by parts, partial fraction decomposition, and numerical integration.