Problem 96 - Calculus Volume 1

Question

In the following exercises, use the identity \(\int_{-A}^{A} f(x) d x=\int_{-A}^{0} f(x) d x+\int_{0}^{A} f(x) d x\) to compute the integrals. \(\int_{1}^{3}(2-x) d x(\text { Hint: Look at the graph of } f .)\)

Step-by-Step Solution

Answer

The definite integral \(\int_{1}^{3} (2-x) \, dx = 0\).

Step 1: Analyze the Identity

The identity \( \int_{-A}^{A} f(x) \, dx = \int_{-A}^{0} f(x) \, dx + \int_{0}^{A} f(x) \, dx \) is useful for splitting integrals over symmetric intervals. However, the given interval is not symmetric about zero. Thus, instead of using this identity, we focus solely on evaluating \(\int_{1}^{3}(2-x) \, dx\), as it is straightforward and doesn't need splitting.

Step 2: Set Up the Integral for Evaluation

First, write the integral that needs evaluation as follows: \( \int_{1}^{3} (2-x) \, dx \). This integral will be evaluated directly since the range is already specified from 1 to 3.

Step 3: Find the Antiderivative

Determine the antiderivative of the function \(2-x\). The antiderivative of \(2\) is \(2x\), and the antiderivative of \(-x\) is \(-\frac{x^2}{2}\). Therefore, the antiderivative is \(F(x) = 2x - \frac{x^2}{2}\).

Step 4: Evaluate the Definite Integral

Substitute the limits of integration into the antiderivative. Calculate \(F(3)\) and \(F(1)\):\[F(3) = 2(3) - \frac{3^2}{2} = 6 - 4.5 = 1.5\]\[F(1) = 2(1) - \frac{1^2}{2} = 2 - 0.5 = 1.5\]Now, find the difference: \(F(3) - F(1) = 1.5 - 1.5 = 0\).

Step 5: Conclusion

The integral over the interval from 1 to 3 is zero, indicating that the areas above and below the x-axis cancel each other out exactly. This results from the linearity and symmetry of the function around the midpoint.

Key Concepts

Antiderivative

The concept of an antiderivative is essential when solving definite integrals. When we talk about an antiderivative, we refer to a function whose derivative gives us the original function we started with.

For instance, to find the definite integral of a function like \(2-x\), we first need to determine its antiderivative. So, we ask ourselves: "What function, when differentiated, results in \(2-x\)?"

• The antiderivative of a constant \(2\) is simply \(2x\), since differentiating \(2x\) yields \(2\).
• For \(-x\), its antiderivative is \(-\frac{x^2}{2}\), as the power rule for derivatives tells us that when we increase the exponent by 1 and divide by the new exponent, differentiating \(-\frac{x^2}{2}\) gives \(-x\).

So, putting these together, the antiderivative of \(2-x\) is the sum: \(F(x) = 2x - \frac{x^2}{2}\). This result sets the stage for evaluating the definite integral, marking the boundary between theory and practical application.

Symmetry in Integrals

Symmetry can simplify the evaluation of definite integrals, especially over symmetric intervals such as \([-A, A]\). This concept suggests that integrals can sometimes be split into parts and evaluated separately if certain symmetries exist.

For our exercise, although the identity \(\int_{-A}^{A} f(x) \, dx = \int_{-A}^{0} f(x) \, dx + \int_{0}^{A} f(x) \, dx\) was cited, it wasn't used because the interval \([1, 3]\) is not symmetric around zero. However, viewing the function's graph or understanding its symmetry can still inform us about potential cancellations.

• If a function is symmetric about the y-axis and the interval is symmetric about the origin,
  the integral over that interval could simplify significantly, as positive and negative contributions cancel.
• In our case, the function \(2-x\) is linear, and the symmetry manifests around the midpoint of the interval itself, resulting in zero after evaluation.

Integral Calculation Steps

When evaluating a definite integral like \(\int_{1}^{3}(2-x) \, dx\), following a systematic approach ensures accuracy. Let's outline the typical calculation steps:

Firstly, ensure the integral setup is correct:

• Identify the function to integrate, \(2-x\), and the interval, \([1,3]\).

Next, find the antiderivative:

• As determined earlier, the antiderivative of \(2-x\) is \(F(x) = 2x - \frac{x^2}{2}\).

Evaluate the antiderivative at the bounds:

• Substitute the upper limit \(3\) into the antiderivative to find \(F(3) = 2(3) - \frac{3^2}{2} = 1.5\).
• Then substitute the lower limit \(1\) giving \(F(1) = 2(1) - \frac{1^2}{2} = 1.5\).

Finally, compute the definite integral by taking the difference between these two results:

• Calculate \(F(3) - F(1) = 1.5 - 1.5\), giving the result as \(0\).

This process reveals how areas under the curve between certain points can cancel each other, explaining why the integral results in zero.