Problem 68 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Verify by direct computation that $$ \int_{0}^{1}\left(1+x-e^{x}\right) d x=\int_{0}^{1} d x+\int_{0}^{1} x d x-\int_{0}^{1} e^{x} d x $$ What properties of the definite integral are demonstrated in this exercise?

Step-by-Step Solution

Answer

In this exercise, we have verified the given equation \( \int_{0}^{1}\left(1+x-e^{x}\right) d x=\int_{0}^{1} d x+\int_{0}^{1} x d x-\int_{0}^{1} e^{x} d x \) by computing the definite integral for each term on both sides. We have demonstrated the properties of definite integral such as linearity of integration and finding antiderivatives. The results of the individual integrals are: \( \int_{0}^{1} d x = 1\), \( \int_{0}^{1} x d x = \frac{1}{2} \), and \( \int_{0}^{1} e^{x} d x = e-1 \). Combining these results, the equation holds true, proving that the given expression is verified.

Step 1: Break down the expression to individual integrals

The given expression is: \( \int_{0}^{1}\left(1+x-e^{x}\right) d x = \int_{0}^{1} d x + \int_{0}^{1} x d x - \int_{0}^{1} e^x d x \). The integral of a sum or difference of functions can be determined as the sum or difference of their individual integrals. Thus, we will compute each individual integral.

Step 2: Compute the integral \(\int_{0}^{1} d x\)

To compute a definite integral, first find the antiderivative of the function and then evaluate it at the limits of integration. The antiderivative of a constant function, in this case, 1, is simply \(x\). So the integral becomes: \[ \int_{0}^{1} d x = (x)\Big|_0^1 = (1-0) = 1 \]

Step 3: Compute the integral \(\int_{0}^{1} x d x\)

The antiderivative of \(x\) is \(\frac{1}{2}x^2\). Therefore, the integral becomes: \[ \int_{0}^{1} x d x = \left(\frac{1}{2}x^{2}\right)\Big|_0^1 = \frac{1}{2}(1^{2} - 0^{2}) = \frac{1}{2} \]

Step 4: Compute the integral \(\int_{0}^{1} e^x d x\)

The antiderivative of \(e^{x}\) is \(e^{x}\). Therefore, the integral becomes: \[ \int_{0}^{1} e^{x} d x = (e^{x})\Big|_0^1 = (e^{1}-e^{0}) = (e-1) \]

Step 5: Combine the results of Steps 2-4

Now, we combine the results of steps 2-4 into the given expression: \[ \int_{0}^{1}\left(1+x-e^{x}\right) d x=\int_{0}^{1} d x+\int_{0}^{1} x d x-\int_{0}^{1} e^{x} d x \] \[(1 + \frac{1}{2}) - (e - 1) = 1 + \frac{1}{2} - e + 1\]

Step 6: Conclusion and properties of the definite integral

Since the equation holds true, we have verified the given expression. The properties of the definite integral demonstrated in this exercise include the linearity of integration (the integral of a sum/difference of functions is the sum/difference of individual integrals of those functions) and the reverse of differentiation (finding antiderivatives).

Key Concepts

Linearity of Integration

The concept of linearity of integration is essential when working with definite integrals. It states that integrating a sum of functions is the same as taking the sum of integrals of each function separately. This property makes calculation simpler by allowing us to break down complex integrals into simpler parts.
For example, if you have a function such as \( f(x) + g(x) - h(x) \), the integration can be separated as:
\[ \int (f(x) + g(x) - h(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx - \int h(x) \, dx \]
This principle greatly simplifies integration tasks by isolating them into manageable chunks. This method was utilized in the original exercise where the function \( 1 + x - e^x \) was split into individual integrals.

Antiderivative

An antiderivative of a function is another function whose derivative is the original function. It is also known as the indefinite integral. Finding the antiderivative is usually the first step in computing a definite integral.
To illustrate, consider the antiderivative of \(1\), which is \(x\). This means that the derivative of \(x\) with respect to \(x\) gives us \(1\). Similarly, the antiderivative of \(x\) is \(\frac{1}{2}x^2\), because differentiating \(\frac{1}{2}x^2\) returns \(x\).
Lastly, the function \(e^x\), which is special due to the fact that it is its own derivative and antiderivative, illustrates this concept well. Thus, \(\int e^x \, dx = e^x + C\), where \(C\) is the constant of integration.

Properties of Definite Integrals

Definite integrals have several properties that are both useful and essential for solving integration problems effectively.
One of the main properties is the ability to evaluate the net area between the function and the horizontal axis over a specific interval \(a,b\). This is done by finding the antiderivative and then applying the limits of integration \(\int_a^b f(x) \, dx = F(b) - F(a)\).
Another vital property is that if the limits of integration are equal, the value of the integral is zero, i.e., \(\int_a^a f(x) \, dx = 0\).
Moreover, scaling or multiplying a function by a constant can be brought outside the integral:

• \(\int_a^b cf(x) \, dx = c \int_a^b f(x) \, dx\)

These properties enable mathematicians and students to easily handle complex integrals by understanding how changes to the function or limits affect the outcome.