Problem 212 - Calculus Volume 1

Question

Use basic integration formulas to compute the following antiderivatives. $$ \int_{0}^{\pi / 2}(x-\sin x) d x $$

Step-by-Step Solution

Answer

\( \frac{\pi^2}{8} - 1 \)

Step 1: Break Down the Integral

The given integral is \( \int_{0}^{\pi / 2}(x-\sin x) \, dx \). It consists of two parts: one involving \( x \) and the other \( \sin x \). We can integrate each part separately due to the linearity of integrals.

Step 2: Integrate the x Term

Integrate the first part, \( \int x \, dx \). The antiderivative of \( x \) with respect to \( x \) is \( \frac{x^2}{2} \). Therefore, \( \int x \, dx = \frac{x^2}{2} + C \), where \( C \) is the constant of integration, which will be determined when evaluating definite integrals.

Step 3: Integrate the sin(x) Term

Next, integrate \( \int \sin x \, dx \). The antiderivative of \( \sin x \) is \( -\cos x \). Thus, \( \int \sin x \, dx = -\cos x + C \).

Step 4: Combine the Antiderivatives

Combine the results from Steps 2 and 3: \( \int (x - \sin x) \, dx = \frac{x^2}{2} - (-\cos x) = \frac{x^2}{2} + \cos x + C \).

Step 5: Evaluate the Definite Integral

Now, evaluate the definite integral from 0 to \( \frac{\pi}{2} \). Substitute the upper limit \( \frac{\pi}{2} \) into the antiderivative: \[ \left[ \frac{x^2}{2} + \cos x \right]_{0}^{\pi / 2} = \left( \frac{(\frac{\pi}{2})^2}{2} + \cos \frac{\pi}{2} \right) - \left( \frac{0^2}{2} + \cos 0 \right) \].

Step 6: Simplify the Result

Evaluate the expression: \( \frac{\pi^2}{8} + 0 - (0 + 1) = \frac{\pi^2}{8} - 1 \). The calculations simplify to \( \frac{\pi^2}{8} - 1 \).

Key Concepts

Antiderivatives

Antiderivatives are an important part of integration and calculus. They are the reverse process of differentiation, finding a function whose derivative gives the original function. Consider the function given by our original problem: \(x - \sin x\). We look for an antiderivative for each part separately.

• For \(x\), the antiderivative is \( \frac{x^2}{2} + C \), because taking the derivative of \( \frac{x^2}{2} \) gives back \(x\).
• For \(-\sin x\), the antiderivative is \(-\cos x + C\), because the derivative of \(-\cos x\) is \(\sin x\). Notice the negative sign!

Breaking down functions into simpler parts and finding each part's antiderivative is a common method in calculus. The antiderivative is not unique due to the arbitrary constant \(C\). Integration focuses on combining these pieces to form the overall antiderivative function.
Understanding antiderivatives helps us solve integrals, especially when dealing with definite integrals, as we'll see later.

Definite Integrals

Definite integrals calculate the area under a curve between two points on the x-axis. Unlike antiderivatives, they have limits of integration, specifying the boundary of the area to calculate. In our exercise, the limits are from \(0\) to \(\frac{\pi}{2}\).

• First, find the indefinite integral or the antiderivative of the function.
• Next, use the limits to evaluate the integral. This involves substituting the upper limit into the antiderivative, then subtracting the result of the lower limit.
• Finally, simplify your expression for the exact numerical result.

For the function \(\int_{0}^{\pi / 2}(x - \sin x) \, dx\), we find its antiderivative first and then apply the limits. Plugging in \(\frac{\pi}{2}\) and \(0\) gives us a result of \(\frac{\pi^2}{8} - 1\).
Definite integration helps in many practical applications, like finding areas, and it's crucial to solve many problems involving accumulated quantities.

Linearity of Integration

The linearity of integration is a useful property that allows us to manipulate integrals easily. It states that you can separate the integral of a sum into the sum of integrals, which can simplify complex expressions. Here's what it involves:

• If you have an integral of a sum, such as \( \int (a(x) + b(x)) \, dx \), you can split this into two separate integrals: \( \int a(x) \, dx + \int b(x) \, dx \).
• It also works for coefficients, where \( \int c \cdot f(x) \, dx = c \int f(x) \, dx \), with \(c\) as a constant.

In our problem, the integral \( \int (x - \sin x) \, dx \) represents a sum of two functions: \(x\) and \(-\sin x\). The linearity property lets us evaluate each part separately, making it straightforward to compute the antiderivative and evaluate the definite integral.
This approach simplifies problems and is a foundational technique in calculus, making integration processes efficient and less time-consuming.