Problem 31 - Calculus of a Single Variable

Question

Evaluating a Definite Integral In Exercises \(21-32\) evaluate the definite integral. $$ \int_{0}^{1 / \sqrt{2}} \frac{\arcsin x}{\sqrt{1-x^{2}}} d x $$

Step-by-Step Solution

Answer

The result of evaluating the integral is \( \frac{\pi}{4} \).

Step 1 - Identify the Derivative

Recognize that this integral represents the derivative of \( \arcsin x\). That is \( \frac{d}{dx} \arcsin x = \frac{1}{\sqrt{1 - x^2}}\). Therefore, the integral can be rewritten as \( \int_{0}^{1 / \sqrt{2}} \frac{d}{dx} (\arcsin x) dx \).

Step 2 - Apply Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that the integral of a derivative over an interval is the difference in the original function evaluated at the endpoints of the interval. Applying this theorem yields \( \arcsin(1 / \sqrt{2}) - \arcsin(0) \).

Step 3 - Evaluate the Function at the Limits

Now, replace x with each limit and subtract: \( \arcsin(1 / \sqrt{2}) - \arcsin(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \).

Key Concepts

Fundamental Theorem of Calculus

Understanding the Fundamental Theorem of Calculus is crucial for evaluating definite integrals, and it can be thought of as the bridge between differential and integral calculus. This theorem has two main parts. The first part tells us that the indefinite integral of a function is related to its antiderivative, and the second part, which applies directly to our exercise, helps us evaluate definite integrals.

To apply the second part of the theorem, take note of the following steps:

• Ensure that the function to be integrated is continuous on the interval from a to b.
• Find an antiderivative of the function. An antiderivative is a function whose derivative is the original function you want to integrate.
• Evaluate the antiderivative at the upper limit (b) and at the lower limit (a) of the definite integral.
• Subtract the value of the antiderivative evaluated at a from the value of the antiderivative evaluated at b to get the result of the integral.

In simple terms, the second part of the Fundamental Theorem of Calculus says that if you have the antiderivative of a function, you can easily find the integral of the function over an interval by evaluating the antiderivative at each endpoint and taking the difference.

Inverse Trigonometric Functions

Inverse trigonometric functions are essential in calculus, especially when dealing with integrals that involve trigonometric expressions. These functions allow us to reverse the action of the usual trigonometric functions (like sine, cosine, and tangent), which means they give the angle whose trigonometric function equals a given value.

For example, where sine gives you the ratio of the opposite side to the hypotenuse in a right triangle, arcsine (written as \(\arcsin\)) will give you the angle whose sine is that ratio. In the given exercise, the presence of \(\arcsin x\) means you're looking for the angle whose sine is x.

Here are some important properties of inverse trigonometric functions to remember:

• Their outputs are angles, typically measured in radians.
• They have specific ranges where they are defined, ensuring they are functions. For instance, the range of \(\arcsin\) is from -\(\pi/2\) to \(+\pi/2\).
• When used in integrals, understanding their derivatives can simplify the problem, as it did in our exercise.

Having a solid grasp of these functions not only aids in integration but also helps to enhance understanding of trigonometry and its applications in calculus.

Definite Integral Calculation

When calculating a definite integral, you are essentially finding the area under the curve of a function between two points on the x-axis. This concept is a fundamental application of integral calculus and is widely used in various fields, such as physics and economics.

The definite integral of a function from a to b is denoted as \(\int_{a}^{b} f(x) dx\), and this problem requires us to use all the tools at our disposal, including the Fundamental Theorem of Calculus and knowledge of derivatives and antiderivatives, as shown in the exercise.

Follow these steps to calculate a definite integral:

• First, ensure the function is integrable over the interval [a, b].
• Find an antiderivative F(x) of the function f(x), which is often the most challenging part.
• Use the limits of integration to evaluate F(x) at b and a.
• Subtract the value of F(a) from F(b) to obtain the result.

The exercise demonstrated these steps by recognizing the integral as the antiderivative of \(\arcsin x\) and applying the Fundamental Theorem of Calculus to evaluate it between the given limits. The result is the area under the curve of the function, conveying important information about the geometric or physical context represented by the function.