Problem 47 - Calculus: One and Several Variables

Question

Derive the following three formulas. Set \(f(x)=x \sin x .\) Find the area between the graph of \(f\) and the \(x\) -axis from \(x=0\) to \(x=\pi\)

Step-by-Step Solution

Answer

The area between the graph of the function \(f(x) = x \sin{x}\) and the x-axis from \(x = 0\) to \(x = \pi\) is \(\pi\) square units. We found this by first deriving \(f'(x) = \sin{x} + x \cos{x}\) using the product rule, and then evaluating the definite integral \(\int_{0}^{\pi} (x \sin{x}) dx\) by applying integration by parts and simplifying the result.

Step 1: Derive \(f'(x)\)

First, we need to find the derivative of the function \(f(x) = x \sin{x}\). We will apply the product rule, which states that \((fg)' = f'g + fg'\), where \(f(x) = x\) and \(g(x) = \sin{x}\). We find the derivatives \(f'(x) = 1\) and \(g'(x) = \cos{x}\): Now, applying the product rule: \[f'(x) = (x \sin{x})' = x' \sin{x} + x \sin'{x} = 1 \cdot \sin{x} + x \cdot \cos{x}\] So, \(f'(x) = \sin{x} + x \cos{x}\).

Step 2: Evaluate the definite integral

Now, let's find the area between the graph of \(f(x) = x \sin{x}\) and the x-axis from \(x = 0\) to \(x = \pi\), by evaluating the definite integral: \[\int_{0}^{\pi} f(x) dx = \int_{0}^{\pi} (x \sin{x}) dx\] To evaluate this integral, we need to perform integration by parts using the following formula: \[\int u dv = uv - \int v du\] Let \(u = x\) and \(\dv{x} = \sin{x} \), then: \[ \du{x} = 1 \] and \( v = -\cos{x}\) Applying the integration by parts formula: \[\int_{0}^{\pi} (x \sin{x}) dx = \left[-x \cos{x}\right]_{0}^{\pi} - \int_{0}^{\pi} (-\cos{x}) dx\] Now, we need to integrate the cosine function over the interval \([0, \pi]\): \[\int_{0}^{\pi} \cos{x} dx = [\sin{x}]_{0}^{\pi}\] Next, we evaluate all the expressions at the boundaries: \[\left[-x \cos{x}\right]_{0}^{\pi} = (-\pi \cos{\pi} - 0) = \pi\] and \[[\sin{x}]_{0}^{\pi} = \sin{\pi} - \sin{0} = 0\] Finally, putting everything together: \[\int_{0}^{\pi} (x \sin{x}) dx = \pi - 0 = \pi\] Hence, the area between the graph of the function \(f(x) = x \sin{x}\) and the x-axis from \(x = 0\) to \(x = \pi\) is \(\pi\) square units.

Key Concepts

Integration by Parts

The method of integration by parts is a crucial technique in calculus, derived from the product rule for differentiation. It enables us to integrate products of two functions that would otherwise be difficult to integrate directly. Conceptually, integration by parts is based on the formula
\[ \int u dv = uv - \int v du \]
where \(u\) and \(dv\) are functions of \(x\). The strategy involves choosing \(u\) and \(dv\) wisely so that the resulting integral \(\int v du\) is simpler than the original.

Applying to our Exercise

In our exercise, we are given the function \(f(x) = x \sin{x}\). Choosing \(u = x\) and \(dv = \sin{x}dx\) simplifies our problem because the derivative of \(x\) is a constant, making the integration process straightforward. After finding \(du\) and integrating \(dv\) to get \(v\), we apply the formula to solve the definite integral, leading us to find the desired area under the curve.

Product Rule for Differentiation

The product rule is an essential differentiation tool that allows us to find the derivative of a product of two functions. If \(f(x)\) and \(g(x)\) are differentiable functions, then the product rule states
\[ (f(x)g(x))' = f'(x)g(x) + f(x)g'(x) \]
This rule simplifies the process of differentiation when faced with products of functions.

In Our Context

For the exercise at hand, we have the function \(f(x) = x \sin{x}\). By applying the product rule, we easily obtain its derivative, \(f'(x) = \sin{x} + x \cos{x}\), which is the first step in finding the exact slope of the function at any point before integrating it to find the area under the curve.

Area Under a Curve

The concept of the area under a curve in calculus represents one of the fundamental applications of the definite integral. The definite integral
\[ \int_{a}^{b} f(x) dx \]
provides a means to calculate the total 'accumulated' area bounded by the curve of the function \(f(x)\), the x-axis, and the vertical lines \(x=a\) and \(x=b\) where \(a\) and \(b\) are the limits of integration.

Exercise Application

The area that we are interested in for our exercise is the region under the curve of \(f(x) = x \sin{x}\) from \(x=0\) to \(x=\pi\). We calculate this area by evaluating the definite integral, which with the aid of integration by parts, yields \(\pi\) square units. This concrete number represents the total area confined within the specified bounds.

Trigonometric Integration

Trigonometric integration involves finding the antiderivative of trigonometric functions. This process is vital for solving problems involving periodic functions or oscillatory behavior. Techniques for integrating trigonometric functions range from basic antiderivatives to more complex methods like trigonometric substitution.

Relation to the Problem

In the exercise, we need to integrate a sine function, one of the basic trigonometric functions. The integral of \(\sin{x}\), which appears in our solution, is \(-\cos{x}\), while the integral of \(\cos{x}\) is \(\sin{x}\). By understanding these fundamental trigonometric integrals, we completed the integration by parts process and, subsequently, discovered the area between the graph of the function and the x-axis.