Question

Sketch the region bounded above by the curve, the x-axis, and x=1, and find the area of the region. Provide the exact form or round answers to the number of places indicated. $$ y=2 x e^{-x} \text { (Approximate answer to four decimal places.) } $$

Short answer

The area of the region is approximately 0.5285.

Step-by-step solution

Identify the boundaries

The problem asks us to find the area of the region bounded by the curve, the x-axis, and the line x = 1. This implies that the region of interest is from x = 0 to x = 1 since the curve starts at x = 0.

Set up the integral

The area under the curve can be found by integrating the function \( y = 2x e^{-x} \) from \( x = 0 \) to \( x = 1 \). We set up the integral as follows: \[ A = \int_{0}^{1} 2x e^{-x} \,dx. \]

Evaluate the integral using integration by parts

To solve the integral \( \int 2x e^{-x} \,dx \), use integration by parts. Select \( u = 2x \) and \( dv = e^{-x} dx \). Then, \( du = 2 dx \) and \( v = -e^{-x} \). The integration by parts formula is: \[ \int u \, dv = uv - \int v \, du. \]

Apply integration by parts

Substitute the values into the integration by parts formula: \[ \int 2x e^{-x} \,dx = -2x e^{-x} \Big|_0^1 + \int 2 e^{-x} \, dx. \] Evaluate this after calculating the right-hand integral.

Solve the remaining integral

The remaining integral \( \int 2e^{-x} \, dx \) results in \(-2e^{-x} \) (consideration of the negative sign due to \( e^{-x} \) derivative). Now evaluate: \[ -2x e^{-x} \Big|_0^1 - 2e^{-x} \Big|_0^1. \]

Evaluate at bounds and calculate the area

Evaluate the expression at the bounds. First, calculate at \( x=1 \): \[ -2(1)e^{-1} - 2e^{-1} = -2e^{-1} - 2e^{-1} = -4e^{-1}. \] Then, at \( x=0 \): \[ -2(0)e^{0} - 2e^{0} = 0 - 2 = -2. \]So, the total area is: \[-4e^{-1} + 2. \]

Approximate the solution

The value \( e^{-1} \) is approximately 0.367879. Substitute to find: \[ A = -4(0.367879) + 2 = -1.471516 + 2 = 0.5285. \] The area of the region is approximately 0.5285.

Key concepts

Integration by Parts

Integration by parts is a powerful technique used to find the integral of the product of two functions. This is particularly useful when dealing with functions that are difficult to integrate directly. The method helps to simplify the problem by breaking it down into more manageable parts.

The core idea is to transform the integral of a product into a simpler form. The formula used is as follows:

• Identify parts of the integral as u and dv.
• Compute the derivative of u (\( du \)) and the integral of dv (\( v \)).
• Substitute into the integration by parts formula: \[\int u \, dv = uv - \int v \, du.\]

In our case, solving the integral \( \int 2x e^{-x} \, dx \) required selecting \( u = 2x \) and \( dv = e^{-x} dx \). Thus \( du = 2 dx \) and \( v = -e^{-x} \), simplifying the process. This transforms the problem into finding \( -2x e^{-x} + \int 2e^{-x} \, dx \), streamlining the solution process.

Area Under the Curve

The area under a curve is a fundamental concept in integral calculus. It represents the accumulation of quantities, which in this context refers to finding the total area between the curve and the x-axis over a specified interval.

To find the area under the curve represented by \( y = 2x e^{-x} \) from \( x = 0 \) to \( x = 1 \), we set up an integral. This is expressed as:

• The integral \( A = \int_{0}^{1} 2x e^{-x} \, dx \) computes the total area.
• The limitations \( x = 0 \) and \( x = 1 \) define the region of interest.
• Solving this integral gives the exact area beneath the curve and the x-axis.

Understanding this helps in visualizing how calculus allows us to measure areas beneath curves, especially when they are irregular or non-linear, beyond simple geometric shapes like rectangles.

Bounded Regions

Bounded regions involve identifying areas that are enclosed by curves and given lines. In calculus, determining these regions often involves analyzing the intersection of different functions and setting limits on integration.

For the given problem, the bounded region is defined by

• The curve \( y = 2x e^{-x} \).
• The x-axis (which acts as a lower boundary).
• The vertical line \( x = 1 \) (which truncates the curve).

By evaluating the bounded region, you precisely define which parts of the function to include or exclude, allowing us to correctly compute areas using integrals. This understanding is crucial for solving problems where multiple boundaries and potential intersections define complex regions, requiring careful setup of integral expressions.