Problem 106 - Precalculus

Question

In calculus, the value of \(F(b)-F(a)\) of a function \(F(x)\) at \(x=a\) and \(x=b\) plays an important role in the calculation of definite integrals. Find the exact value of \(F(b)-F(a)\). $$F(x)=\sin ^{3} x, a=0, b=\frac{\pi}{4}$$

Step-by-Step Solution

Answer

\( F(b) - F(a) = \frac{\sqrt{2}}{4} \)

Step 1: Evaluate the Function at b

First, we need to evaluate the function \( F(x) = \sin^3 x \) at \( x = b \), where \( b = \frac{\pi}{4} \). Calculate \( F\left( \frac{\pi}{4} \right) = \left( \sin \frac{\pi}{4} \right)^3 \). The value of \( \sin \frac{\pi}{4} \) is \( \frac{\sqrt{2}}{2} \), so \( F\left( \frac{\pi}{4} \right) = \left( \frac{\sqrt{2}}{2} \right)^3 = \frac{2\sqrt{2}}{8} = \frac{\sqrt{2}}{4} \).

Step 2: Evaluate the Function at a

Next, evaluate the function \( F(x) = \sin^3 x \) at \( x = a \), where \( a = 0 \). Calculate \( F(0) = (\sin 0)^3 = 0^3 = 0 \).

Step 3: Calculate the Difference

Finally, calculate \( F(b) - F(a) \) using the values obtained. We have \( F\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{4} \) and \( F(0) = 0 \). Therefore, \( F(b) - F(a) = \frac{\sqrt{2}}{4} - 0 = \frac{\sqrt{2}}{4} \).

Key Concepts

Function Evaluation

Function evaluation is a core concept in mathematics where we determine the output of a function for given input values. In the context of definite integrals, it involves calculating the function values at the boundaries of the interval we're interested in. This is essential because it's part of understanding the Fundamental Theorem of Calculus.To perform function evaluation, we replace the variable in the function with the given number, also known as the argument. For example, given that the function is defined as \(F(x) = \sin^3 x\), and our task is to evaluate this at specific points such as \(x = 0\) and \(x = \frac{\pi}{4}\), we substitute these values into the function.

• For \(x = \frac{\pi}{4}\), we substitute into \(F(x)\) to get \(F\left( \frac{\pi}{4} \right) = \left( \sin \frac{\pi}{4} \right)^3\).
• Similarly, for \(x = 0\), we plug it into the function to get \(F(0) = \sin^3 (0) = 0\).

Replacing the input with known values allows us to compute the numerical outcome of the function, which is crucial when determining definite integrals.

Trigonometric Functions

Trigonometric functions are fundamental in mathematics, involving the relationships between angles and sides of a triangle. They are also pivotal in calculus, especially when evaluating integrals.The sine function, \(\sin x\), is one of the basic trigonometric functions that show how much one of the triangle's sides shrinks or stretches. When you cube this function, as in our example \(\sin^3 x\), it alters the amplitude further, affecting its graph and values.

• The sine function at \(x = \frac{\pi}{4}\) simplifies to \(\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}\).
• By cubing this value, we calculate \(\left( \frac{\sqrt{2}}{2} \right)^3\), which is essential in evaluating the function smoothly through this specific angle.

Trigonometric identities and simplifications often help proceed with more complex calculus problems, making these functions a staple in math education.

Calculus

Calculus is a powerful mathematical tool used to study changes and quantities. It consists of two main parts: differential calculus, which deals with rates of change, and integral calculus, which involves finding the quantity where the change is known.In our task, we are dealing with definite integrals, particularly finding the change in a function over an interval \([a, b]\). The definite integral is the signed area under the curve of a function, providing valuable insights into the accumulation of quantities.To solve a problem using definite integrals, we rely on the Fundamental Theorem of Calculus:\[ F(b) - F(a) \]This gives the net change over the interval \([a, b]\) for the antiderivative \(F(x)\). By calculating \(F(\frac{\pi}{4}) - F(0)\), we understand the exact change in the function \(\sin^3 x\) over the interval from 0 to \(\frac{\pi}{4}\).Understanding definite integrals helps you to solve a variety of real-world problems, such as calculating distances, areas, and even predicting future outcomes based on current rates of change.