91Ó°ÊÓ

Calculating the area of a region bounded by curves is a fundamental concept in calculus. In our case, we are considering the area bounded by the curve \(y = \tan^{-1}x\), the x-axis (\(y=0\)), and the vertical line at \(x = 1\). Finding this area involves setting up an integral.
The integral for the area, \(A\), is set up as \(\int_{0}^{1} \tan^{-1} x \, dx\). This integral computes the accumulated area under the curve from \(x = 0\) to \(x = 1\).
To solve:

• We use integration by parts, a technique useful for integrating products of functions.
• Choose \(u = \tan^{-1} x\) and \(dv = dx\). Then, \(du = \frac{1}{1+x^2} dx\) and \(v = x\).
• The integration by parts formula \(\int u \, dv = uv - \int v \, du\). This helps break down complex integrals.

Integrating step-by-step, substituting back into the formula, and evaluating within bounds gives us the area, \(\frac{\pi}{4} - \frac{1}{2} \ln(2)\). This formula is derived from evaluating the integral within the specified limits.

A fascinating application of definite integrals is the calculation of the volume of a solid of revolution. In our scenario, we rotate the region around the y-axis to form a 3D shape.
We use the shell method, which involves the formula \(V = 2\pi \int_{0}^{1} x (\tan^{-1} x) \, dx\). This formula calculates the volume by integrating cylindrical shells formed by vertical slices of the region
The process includes:

• Choosing \(u = \tan^{-1} x\) and \(dv = x \, dx\). Thus, \(du = \frac{1}{1+x^2} \, dx\) and \(v = \frac{x^2}{2}\).
• Using integration by parts helps divide the integral into manageable parts.

Using the integration by parts formula gives \(2\pi \left(\frac{x^2}{2} \tan^{-1} x - \int \frac{x^3}{2(1+x^2)} \, dx\right)\).
After simplifying and solving, calculating within the limits 0 to 1 gives the volume as \(\frac{\pi^2}{8} - \frac{\pi}{4} \ln 2\). This result provides the exact volume generated by rotating the region about the y-axis.

Integration by parts is a powerful technique used to tackle integrals of products of functions. This formula is based on the product rule for differentiation and is essential in situations where simpler integral methods fall short.
The formula is expressed as \(\int u \, dv = uv - \int v \, du\). It requires splitting an integral into two components: \(u\) and \(dv\).
Steps include:

• Select \(u\) and \(dv\) from the product such that \(du\) and \(v\) are easily computable. The choice of \(u\) is critical; a good choice simplifies the integration process.
• The product \(uv\) is immediately evaluated.
• Compute the integral \(\int v \, du\): the substitution of manageable parts into simpler integrals is key.

This technique is employed twice in the exercise: first, to find the area under \(y = \tan^{-1}x\) and then to calculate the volume of revolution using the shell method. Mastery of this technique expands your integral-solving toolbox significantly.