Problem 50 - Calculus: One and Several Variables

Question

Find the area between the graph of \(f\) and the \(x\) axis. $$f(x)=x e^{-2 x}, \quad x \in[0,2]$$

Step-by-Step Solution

Answer

The area between the graph of \(f(x) = x e^{-2x}\) and the x-axis on the interval [0, 2] is \(-\frac{3}{4} e^{-4} - \frac{1}{4}\).

Step 1: Integrate the function

First, we need to find the integral of \(f(x) = x e^{-2x}\) with respect to x. To do this, we will use integration by parts: Let: \(u = x\) and \(dv = e^{-2x}dx\) Now, differentiate u and integrate dv: \(du = dx\) \(v = -\frac{1}{2}e^{-2x}\) Using the integration by parts formula, we get: \[\int x e^{-2x}dx = uv - \int v du\]

Step 2: Apply the integration by parts formula

Now we apply the integration by parts formula by plugging in the variables we found earlier: \[\int x e^{-2x} dx = -\frac{1}{2}xe^{-2x} - \int -\frac{1}{2}e^{-2x} dx\] Simplify and integrate: \[\int x e^{-2x} dx = -\frac{1}{2}xe^{-2x} + \frac{1}{4}e^{-2x} + C\]

Step 3: Apply the fundamental theorem of calculus

Now that we have found the indefinite integral of the function, we can find the area between the graph of the function and the x-axis on the interval [0, 2] by applying the fundamental theorem of calculus: \[Area = \int_{0}^{2} x e^{-2x} dx = \left[-\frac{1}{2}xe^{-2x} + \frac{1}{4}e^{-2x}\right]_{0}^{2}\]

Step 4: Calculate the area

Plug the limits of integration into our result to find the area: \[Area = \left[-\frac{1}{2}(2)e^{-2(2)} + \frac{1}{4} e^{-2(2)}\right] - \left[-\frac{1}{2}(0)e^{-2(0)} + \frac{1}{4}e^{-2(0)}\right]\] Now, simplify and calculate: \[Area = \left[-e^{-4} + \frac{1}{4} e^{-4}\right] - \left[0 + \frac{1}{4}\right]\] \[Area = -\frac{3}{4} e^{-4} - \frac{1}{4}\] Thus, the area between the graph of the function \(f(x) = xe^{-2x}\) and the x-axis on the interval [0, 2] is \(-\frac{3}{4} e^{-4} - \frac{1}{4}\).

Key Concepts

Definite Integral

A definite integral is a fundamental tool in calculus that allows you to compute the total accumulation of a quantity, such as area, under a curve over a specific interval. Unlike an indefinite integral, which represents a family of functions, a definite integral yields a numerical value. This value corresponds to the area between the curve of the function and the x-axis on the interval from point A to point B.

When calculating a definite integral, you write it as follows:

• \( \int_{a}^{b} f(x) \, dx \)

Here, \( a \) and \( b \) are the limits of integration, which define the interval, while \( f(x) \) is the function you are integrating. You calculate the definite integral by finding the indefinite integral, and then evaluating it at the upper limit and subtracting the result of evaluating it at the lower limit.

For example, in the original problem, after finding the indefinite integral of \( x e^{-2x} \), we evaluated this function from 0 to 2 using the limits to compute the actual area under the curve over that interval.

Fundamental Theorem of Calculus

The fundamental theorem of calculus bridges the two main concepts of calculus: differentiation and integration. Essentially, it states that differentiation and integration are inverse processes. The theorem has two parts, and together, they help connect the concept of an antiderivative with the evaluation of a definite integral.

The first part of the theorem states that if you have a function that is continuous over an interval, the integral of this function over that interval can be found by taking the difference of an antiderivative evaluated at the endpoints of the interval. In formula terms, it looks like this:

• \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \)

Here, \( F \) is any antiderivative of \( f \). The second part concerns itself with the derivative of this integral function butin the context of the problem provided, the first part is particularly useful.

Applying this theorem in our problem let us determine the area under the curve of \( f(x) = x e^{-2x} \) over the interval [0, 2] by evaluating the found antiderivative at these endpoints. This process simplified the calculation of the area and demonstrated the theorem in action.

Area Under Curve

Finding the area under a curve is a common application of definite integrals. In many real-world scenarios, such as physics and economics, this helps in understanding accumulated quantities. For any function \( f(x) \) that is greater than or equal to zero over an interval \([a, b]\), the definite integral \( \int_{a}^{b} f(x) \, dx \) gives the exact area under the curve from \( x = a \) to \( x = b \).

The process involves several steps:

• Identify the function and interval over which you want the area.
• Calculate the indefinite integral (antiderivative) of the function.
• Apply the fundamental theorem of calculus to evaluate the definite integral, determining the area.

In the problem you are exploring, the task was to calculate the area under the curve defined by \( f(x) = x e^{-2x} \) on the interval from 0 to 2. Through integration by parts and definite integration from the limits 0 to 2, the area was computed accurately, illustrating how calculus tools are used to solve such practical problems. This approach allows for accurate computation of areas even for curves that may not be graphically simple.