Question

(a) verify the integration formula, and (b) use it to find the area of the region. $$ \int \frac{\arctan x}{x^{2}} d x=\ln x-\frac{1}{2} \ln \left(1+x^{2}\right)-\frac{\arctan x}{x}+C $$

Short answer

The integral formula \(\int \frac{\arctan x}{x^{2}} dx = \ln x -\frac{1}{2} \ln\left(1+x^{2}\right)-\frac{\arctan x}{x}+C\) holds true upon differentiation. The area under the curve between a to b is given by \(\int_a^b \frac{\arctan x}{x^{2}} dx = (\ln b -\frac{1}{2} \ln\left(1+b^{2}\right)-\frac{\arctan b}{b}) - (\ln a -\frac{1}{2} \ln\left(1+a^{2}\right)-\frac{\arctan a}{a})\).

Step-by-step solution

Verify the integration formula

Take the derivative of the right hand side of the formula and see if it matches the given integrand. To differentiate \(ln x\), use the formula: \[\frac{d}{dx} ln x = \frac{1}{x}\]. Also, you need to use the chain rule for differentiating \(ln(1+x^2)\) and \(\arctan x\). Note, \[\frac{d}{dx}\ln(1+x^2) = \frac{2x}{1+x^2}\] and \[\frac{d}{dx} (\arctan x) =\frac{1}{1+x^2}\].

Simplify the obtained derivative

Simplify the derivative based on fundamental calculus rules and validate that the derived expression equals the integrand \(\frac{\arctan x}{x^{2}}\).

Use the verified integral to find the area

After we have confirmed that the integration formula \(\int \frac{\arctan x}{x^{2}} dx = \ln x -\frac{1}{2} \ln\left(1+x^{2}\right)-\frac{\arctan x}{x}+C\) holds true, we can use it to compute the area under the curve from a to b. This is given by \(\int_a^b \frac{\arctan x}{x^{2}} dx = (\ln b -\frac{1}{2} \ln\left(1+b^{2}\right)-\frac{\arctan b}{b}) - (\ln a -\frac{1}{2} \ln\left(1+a^{2}\right)-\frac{\arctan a}{a})\). Substitute the end points of the region to find the area.

Key concepts

Integral Calculus

Integral calculus is a fundamental part of mathematics that deals with the accumulation of quantities, such as areas under curves. It contrasts with differential calculus, which involves rates of change and slopes of curves. The process of integration, which is the core operation in integral calculus, often necessitates finding an antiderivative.

In the exercise provided, the integration formula given as \[ \int \frac{\arctan x}{x^{2}} dx = \ln x -\frac{1}{2} \ln(1+x^{2})-\frac{\arctan x}{x}+C \] needs verification. Here, to verify the formula, we differentiate the right-hand side because integration and differentiation are inverse processes. If differentiation of the right-hand side equals the original integrand, the integration formula is correct. To achieve this verification, we use known differentiation formulas for natural logarithms and inverse trigonometric functions, ensuring that students understand the underlying principles guiding these operations for proper application.

Furthermore, the process emphasizes the importance of understanding these mathematical tools, allowing for extensions beyond simple verification to practical applications such as calculating areas.

Area Under a Curve

The concept of 'area under a curve' is an application of integral calculus where integration is used to find the space enclosed by a graph of a function and the x-axis, across a particular interval. This measure is useful in various fields, including physics, economics, and engineering, where it can represent things like distance traveled over time or the total cost over a range of production levels.

In the context of the problem, once the integration formula is verified, it can be utilized to calculate the exact area under the curve represented by \( \frac{\arctan x}{x^{2}} \). By evaluating the definite integral from a to b, which means considering the area between these two points on the x-axis, students can determine the actual area under the curve. This is an essential skill as it demonstrates how mathematical procedures translate into tangible results and reinforces their grasp of the direct link between abstract formulas and real-world quantities.

Differentiation and Integration

Differentiation and integration are two core operations of calculus with a deep interrelationship. Differentiation measures the instantaneous rate of change of a function, like how fast an object is moving at one moment. Integration, in contrast, aggregates these changes to give a whole picture, such as the total distance an object has traveled over time.

The problem explicitly utilizes this relationship for formula verification, with differentiation acting as the checking mechanism. To solidify students' understanding, we considered the derivatives of natural logarithmic functions and the arctangent function, utilizing rules like the chain rule to navigate complex functions. This exercise serves as a reinforcement of the underlying principle that integration and differentiation are interconnected, and mastering both is essential to excelling in calculus. Students are encouraged to practice transforming between forms (differentiating an integral expression and vice versa) as this underpins a significant portion of problem-solving in higher mathematics.