Problem 22 - Advanced Engineering Mathematics

Question

Evaluate the double integral over the region \(R\) that is bounded by the graphs of the given equations. Choose the most convenient order of integration. $$ \iint_{R} x d A ; y=\tan ^{-1} x, y=0, x=1 $$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{1}{4}(\pi - \ln(2)) \).

Step 1: Understand the Region R

Firstly, let's determine the region do be integrated over. We have the region bounded by \( y = \tan^{-1}(x) \), \( y = 0 \), and \( x = 1 \). The curve \( y = \tan^{-1}(x) \) represents a function from \( x = 0 \) to \( x = 1 \), because as \( x \to \infty \), \( \tan^{-1}(x) \to \frac{\pi}{2} \). Since within these bounds \( y \) varies from 0 to \( \tan^{-1}(x) \), the region \( R \) is defined in the first quadrant where \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq \tan^{-1}(x) \).

Step 2: Write the Integral

Given the region and the function, the integral can be represented in two orders: \((x, y)\) or \((y, x)\). Since the limits of \( y \) are more complex due to the \( \tan^{-1}(x) \), we'll choose to integrate in the order of \( dy \) first and \( dx \) next. The integral becomes: \[ \int_{x=0}^{1} \int_{y=0}^{\tan^{-1}(x)} x \, dy \, dx \]

Step 3: Integrate with respect to y

Next, we integrate the inner integral with respect to \( y \). Calculating \( \int_{y=0}^{\tan^{-1}(x)} x \, dy \), since \( x \) is constant in terms of \( y \), it pulls out of the integral: \[ x \int_{y=0}^{\tan^{-1}(x)} \, dy = x \cdot [ y ]_{0}^{\tan^{-1}(x)} = x (\tan^{-1}(x) - 0) = x \tan^{-1}(x) \]

Step 4: Integrate with respect to x

Now, we integrate with respect to \( x \) from 0 to 1: \[ \int_{x=0}^{1} x \tan^{-1}(x) \, dx \] This is an integration by parts problem. Here, let \( u = \tan^{-1}(x) \) and \( dv = x \, dx \). Then, we would need to find \( du \) and \( v \). \( du = \frac{1}{1+x^2} dx \) and \( v = \frac{x^2}{2} \).

Step 5: Apply Integration by Parts

Integration by parts formula is: \[ \int u \, dv = uv - \int v \, du \] where \( u = \tan^{-1}(x), dv = x \, dx \) implying \( du = \frac{1}{1+x^2} \, dx \) and \( v = \frac{x^2}{2} \), then:\[ \left[ \frac{x^2}{2} \tan^{-1}(x) \right]_0^1 - \int_0^1 \left[ \frac{x^2}{2} \cdot \frac{1}{1+x^2} \right] dx \]Evaluate at bounds:\[ \left[ \frac{1}{2} \cdot \tan^{-1}(1) - 0 \right] - \int_0^1 \frac{x^2}{2(1+x^2)} \, dx \]

Step 6: Solve the Remaining Integral

The remaining integral \( \int_0^1 \frac{x^2}{2(1+x^2)} \, dx \) can be approached through substitution. Let \( u = 1 + x^2 \), then \( du = 2x \, dx \). Substitute and simplify, setting limits when \( x = 0, u = 1 \) and when \( x = 1, u = 2 \).Rewriting the integral for simplification: \[ \int_0^1 \frac{x^2}{2(1+x^2)} \, dx = \frac{1}{2} \left[ \frac{1}{2}\ln(1+x^2) \right]_0^1 \] Evaluate this integral to get \( \frac{1}{4} \left( \ln(2) \right) \).

Step 7: Evaluate and Combine Results

Insert the solved value of the integral and constant terms in Step 5 and 6: \[ \frac{1}{4} \cdot \pi - \frac{1}{4} \ln(2) \] This yields the final result of the evaluated double integral over region \( R \).

Key Concepts

Integration by Parts

The technique of integration by parts is pivotal in solving integrals that result in products, such as \( \int x \tan^{-1}(x) \, dx \). This method is derived from the product rule of differentiation, inversely applied to integration. The formula is:\[ \int u \, dv = uv - \int v \, du \]Where:

• \( u \) is a function which becomes simpler when differentiated.
• \( dv \) is a function that is easy to integrate.

Applying this method involves choosing \( u = \tan^{-1}(x) \) and \( dv = x \, dx \). Therefore:

• \( du = \frac{1}{1+x^2} \, dx \)
• \( v = \frac{x^2}{2} \)

Executing the values within the integration by parts formula simplifies the original integral into manageable components. This method is a useful tool for tackling integration problems involving functions multiplied by other algebraic functions.

Arctangent Function

The arctangent function, denoted as \( \tan^{-1}(x) \), is an inverse trigonometric function. Its purpose is to return the angle whose tangent is \( x \). This makes it quite useful in varying applications, particularly in calculus.Essential properties of \( \tan^{-1}(x) \):

• The domain of \( \tan^{-1}(x) \) is all real numbers.
• The range is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).
• \( \lim_{x \to \infty} \tan^{-1}(x) = \frac{\pi}{2} \).
• \( \lim_{x \to -\infty} \tan^{-1}(x) = -\frac{\pi}{2} \).

In the context of the double integral problem, the arctangent function helps define the bounds of the region \( R \). It acts as an upper boundary for \( y \) and is key when integrating as the limits change along the path defined by \( \tan^{-1}(x) \).