Problem 12 - Calculus: Early Transcendentals

Question

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. \( \displaystyle R(y) = \int^2_y t^3 \sin t \,dt \)

Step-by-Step Solution

Answer

The derivative is \( R'(y) = -y^3 \sin y \).

Step 1: Understand the Function and the Theorem

The function given is \( R(y) = \int^2_y t^3 \sin t \, dt \). The Fundamental Theorem of Calculus Part 1 states that if \( F(x) = \int_a^x f(t) \, dt \), then \( F'(x) = f(x) \). Our aim is to find the derivative of \( R(y) \).

Step 2: Invert the Limits of Integration

The integral \( \int^2_y \) has a lower limit that is a variable \( y \). To apply the Fundamental Theorem of Calculus directly, we invert the integration limits. This changes the sign: \( R(y) = -\int^y_2 t^3 \sin t \, dt \).

Step 3: Apply the Fundamental Theorem of Calculus

Now use the theorem that says if \( F(y) = \int_a^y f(t) \, dt \), then \( F'(y) = f(y) \). For \( R(y) = -\int^y_2 t^3 \sin t \, dt \), we have \( R'(y) = -[y^3 \sin y] \).

Step 4: Simplify the Derivative

The derivative from Step 3 is \( R'(y) = -y^3 \sin y \). This is the derivative of the function \( R(y) \).

Key Concepts

Integration by Parts

Integration by parts is a powerful integration technique that stems from the product rule of differentiation. It is particularly useful when integrating the product of two functions. The formula for integration by parts is given by:\[ \int u \, dv = uv - \int v \, du \]Here, you identify parts of the integrand as one function, \( u \), and its differential, \( dv \), then differentiate \( u \) to get \( du \) and integrate \( dv \) to get \( v \). This technique is most effective when the differentiation simplifies one part of the integrand or when the resulting integral is easier to evaluate.

• Pick \( u \) and \( dv \) wisely to ease integration.
• Remember, the product \( uv \) is evaluated before the second integral.

Integration by parts can be applied repeatedly or in conjunction with other techniques to solve complex integrals.

Derivative

The derivative is a fundamental concept in calculus that represents the rate of change of a function. In simple terms, it tells us how a function is changing at any point. The process of finding a derivative is called differentiation.

• For a function \( f(x) \), the derivative \( f'(x) \) is defined as \( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \).
• The derivative gives the slope of the tangent line to the curve at any point.

In the problem, we were tasked with finding the derivative of the function \( R(y) = \int_2^y t^3 \sin t \, dt \). Using the Fundamental Theorem of Calculus, this derivative simplifies the problem of finding how the accumulated area changes as \( y \) changes. Recognizing the derivative is crucial for understanding many physical phenomena, from velocity in physics to marginal cost in economics, because it quantifies how things change.

Integration Techniques

Various integration techniques can be applied to solve complex integrals beyond basic antiderivatives. Knowing when and how to use these methods helps solve problems that seem challenging at first.1. **Basic Integration:** Find antiderivatives of simple functions.2. **Integration by Parts:** Useful for integrating the product of two functions.3. **Substitution:** Simplifies integrals by changing variables, akin to the chain rule in derivatives.4. **Trigonometric Identities:** Use trigonometric identities to simplify integrals involving trigonometric functions.5. **Partial Fractions:** Break down rational functions into simpler fractions for easier integration.By employing these techniques, you can tackle integrals that may initially appear intimidating. Knowing the appropriate technique for the problem is part of the art of integration. Practice is essential to develop an instinct on which method to choose, especially in complex functions like \( t^3 \sin t \). Integration is not just about finding areas under curves but also applying these techniques to real-world problems.