91Ó°ÊÓ

Integration by parts is a technique used to solve integrals, particularly when the integral is a product of two functions. It's similar to the product rule for differentiation. This method is essential when you can't directly integrate a function easily. The formula for integration by parts is given by:\[\int u \, dv = uv - \int v \, du\]To use integration by parts, you need to choose two parts of the integrand, identifying one as \(u\) and the other as \(dv\). In this context:

Choose \(u = \tan^{-1}(x)\), and find its derivative \(du = \frac{1}{1+x^2} \, dx\).

Choose \(dv = dx\), which leads to \(v = x\) after integration.

This choice allows us to transform the original integral into a simpler form. Once identified, apply the formula to break down the integral:

• The integral transforms into \(x \tan^{-1}(x) - \int \frac{x}{1+x^2} \, dx\).

Notice how the second integral simplifies the problem by using substitution. By solving this, we step closer to finding the area under the curve.

Finding the area under a curve is a common problem in calculus. It involves integrating the function between the given interval. In this exercise, we are looking at the area beneath the curve \(y = \tan^{-1}(x)\) from \(x = 0\) to \(x = 1\). The general integral for this area is:\[A = \int_{0}^{1} \tan^{-1}(x) \, dx\]The area involves calculating this integral. After setting up the integral, integration by parts is applied, resulting in:

• \(x \tan^{-1}(x)\) is evaluated from 0 to 1.
• An additional integral \(\int \frac{x}{1+x^2} \, dx\) is solved through substitution, giving \(\frac{1}{2} \ln(1+x^2)\).

The final step is to evaluate the definite integral at the boundaries. This would be:

• Substituting \(x=1\) gives \(\frac{\pi}{4} - \frac{1}{2} \ln(2)\).
• Subtracting the result of \(x=0\), which is zero since \(\ln(1) = 0\).

Hence, the area under the curve is determined as \(\frac{\pi}{4} - \frac{1}{2} \ln(2)\), successfully capturing the space beneath the curve from \(x=0\) to \(x=1\).

The volume of a solid of revolution is the total three-dimensional space enclosed within a solid obtained by spinning a region around an axis. For this, the washer method is frequently used. Essentially, it involves slicing the solid perpendicular to the axis of revolution, forming circular disks with a hole, or washers.In this particular problem, we revolve the area bounded by \(y = \tan^{-1} x\), \(y = 0\), and \(x = 1\) around the y-axis. The integral for the volume using the washer method is:\[V = \pi \int_{0}^{\pi/4} (1 - (\tan(y))^2) \, dy\]Here, the integral bounds are from 0 to \(\pi/4\), considering the region between \(y = \tan^{-1}(0)\) and \(y = \tan^{-1}(1)\). The equation \(x = \tan(y)\) helps re-parameterize the bounds from \(x\) to \(y\).Finally, evaluating this integral gives:

• The expression evaluates to \(V = \pi(\frac{\pi}{4} - \frac{1}{3})\).
• Ultimately, simplify to reach the precise volume of the solid: \(\frac{\pi^2}{4} - \frac{\pi}{3}\).

This resulting value represents the three-dimensional volume enclosed when the specified area is revolved around the y-axis.