Problem 14 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Evaluate \(\int_{0}^{0.5} \arcsin x d x\)

Step-by-Step Solution

Answer

-\(\pi/12\)

Step 1: Understanding the integral

The exercise asks to evaluate the integral of the function \(f(x) = \arcsin(x)\) from 0 to 0.5. That is, we're looking for the area under the graph of \(f(x)\) between those two points.

Step 2: Applying Bernoulli's rule

This can be calculated by integrating by parts using the Bernoulli's rule: \(\int_{a}^{b} u(x) v(x) d x=[u(x) v(x)]_{a}^{b}-\int_{a}^{b} v(x) u'(x) d x\). Here, let's use \(u(x) = x\) and \(v(x) = \arcsin(x)\). Then, \(du = dx\) and \(v' = 1/\sqrt{1-x^2}\). Substitute these functions and their derivatives into Bernoulli's rule.

Step 3: Calculation

\(\int_{0}^{0.5} x \frac{1}{\sqrt{1-x^2}} dx - x\arcsin(x) |_{0}^{0.5} = 0.5 \arcsin(0.5) - \int_{0}^{0.5} \frac{1}{\sqrt{1-x^2}} dx. Then solve \( \int_{0}^{0.5} \frac{1}{\sqrt{1-x^2}} dx \), which results in \( [\arcsin(x)]_{0}^{0.5} = \arcsin(0.5) - \arcsin(0) = \pi/6\). Substitute this back into the previous equation \).

Step 4: Final Solution

After calculating, substitute the above result into the equation: \(0.5 \arcsin(0.5) - \pi/6 = \frac{\pi}{12} - \frac{\pi}{6} = -\frac{\pi}{12}\).

Key Concepts

Integration by Parts

Integration by Parts is a powerful technique used to compute integrals where the standard approach might be challenging. This method is inspired by the product rule for differentiation. When you come across a function that's a product of two different types of functions, such as algebraic with trigonometric, or logarithmic with exponential, Integration by Parts is often a great choice.

The formula for Integration by Parts is:\[ \int u(x) \cdot v'(x) \, dx = \left[ u(x) \cdot v(x) \right]_a^b - \int v(x) \cdot u'(x) \, dx \]
This method requires selecting one part of the integrand as \( u(x) \) and another as \( v'(x) \). You then compute the derivatives and integrals of these functions accordingly.

• Select \( u(x) \) such that its derivative becomes simpler.
• Select \( v'(x) \) such that its integral is easily computable.
• Apply the formula and simplify carefully, considering the limits when dealing with a definite integral.

In the given problem, \( u(x) \) was chosen as \( x \) and \( v(x) \) as \( \arcsin(x) \). Professional judgment in such selections eases the path to solving tricky integrals.

Inverse Trigonometric Functions

Inverse trigonometric functions, like \( \arcsin(x) \), reverse the effects of standard trigonometric functions. While trigonometric functions generally take angle measures and return ratios, their inverses do the opposite: take a ratio and provide an angle. For example, \( \arcsin(x) \) answers the question, "What angle \( \theta \) from \( -\pi/2 \) to \( \pi/2 \) has a sine of \( x \)?"

Their presence in integrals calls for specific methods due to their unique properties:

• Derivative of \( \arcsin(x) \) is \( \frac{1}{\sqrt{1-x^2}} \).
• Useful in reversing situations where angles need extracting from known sine values.
• Have specific domain and range constraints (e.g., \( \arcsin(x) \) is only defined for \( x \) between -1 and 1).

In our exercise, integrating \( \arcsin(x) \) involves knowing its derivative, which simplifies one part of our computation. These functions hold an elegance with their geometric and analytic shapes, making them useful in a wide array of applications from physics to engineering.

Area under the Curve

The concept of finding the "Area under the Curve" represents one of the foundational ideas within calculus. It essentially refers to integrating a function between given limits to calculate the size of the region between the curve and the x-axis.

This area provides valuable physical quantities such as distance, probability, and more, depending on the context of the function:

• For a definite integral with limits \( a \) to \( b \), it computes \( \int_a^b f(x) \, dx \).
• Ultimately, it measures the cumulative total between curve points across an interval on the x-axis.
• The result can be positive, negative, or zero, depending on whether the function lies above or below the x-axis.

In the given problem, the task was to find the area under the curve of the \( \arcsin(x) \) function from 0 to 0.5. Understanding that the integral of a function gives its accumulated 'area' makes this process crucial for interpreting real-world data modeled by mathematical functions.