Problem 19 - Applied Calculus

Question

Find the integrals in problems. Check your answers by differentiation. $$ \int \sin (3-t) d t $$

Step-by-Step Solution

Answer

The integral is \( \cos(3-t) + C \).

Step 1: Identify the Inner Function

The integrand is \( \sin(3-t) \). The inner function here is \( 3-t \). This step involves recognizing that we can use substitution to simplify the integration process.

Step 2: Perform Substitution

Let \( u = 3-t \). Then, differentiate to find \( du \): \( du = -dt \) or \( dt = -du \). Substitute these expressions into the integral: \[ -\int \sin(u) du. \]

Step 3: Integrate with Respect to the New Variable

The integral becomes:\[ -\int \sin(u) du = \cos(u) + C,\]where \( C \) is the constant of integration. The integral of \( \sin(u) \) is \( -\cos(u) \). By factoring out the negative sign we get \( \cos(u) \).

Step 4: Reverse the Substitution

Substitute back the value of \( u = 3-t \) to obtain the integral in terms of \( t \): \[ \cos(3-t) + C.\]

Step 5: Verify the Solution by Differentiation

Differentiate the answer \( \cos(3-t) + C \) with respect to \( t \):\[ \frac{d}{dt}[\cos(3-t) + C] = \sin(3-t). \]The derivative \( \frac{d}{dt}[\cos(3-t)] = \sin(3-t) \), matching the original integrand, thus confirming the solution is correct.

Key Concepts

Integration

Integration is a fundamental concept in calculus, often described as the inverse process of differentiation. It allows us to find the accumulation of quantities, such as areas under curves. When integrating a function, you essentially "add up" infinitely small sections of the function to find a total value.
Integration can be viewed as finding the antiderivative of a function. This means identifying a function whose derivative is equal to the integrand (the function you're integrating).

• An integral may have a constant term (denoted as +C), representing an infinite family of functions differing by a constant.
• The most common methods of integration include substitution, integration by parts, and trigonometric identities.
• In definite integrals, limits are applied to calculate a precise area, whereas in indefinite integrals, you find the general form of the antiderivative.

Substitution Method

The substitution method is a powerful technique used to simplify an integral, making it easier to solve. It involves changing the variable of integration to convert a difficult integral into a more manageable one.
The key steps in substitution are:

• Identify the inner function: Look for a function within the integrand that can be substituted with a single variable (often called 'u').
• Set the substitution: Let 'u' be equal to the identified inner function. This will translate the integral into a new variable.
• Derive 'du': Differentiate 'u' to find 'du', ensuring you replace 'dt' or the original differential as needed.
• Rewrite the integral: Substitute 'u' and 'du' back into the integral and solve.
• Reverse the substitution: Once the integral is solved, substitute back the original variables to finish.

Using substitution makes it easier to tackle integrals with nested functions, as seen in integrating \(\int \sin(3-t) dt\).

Differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate at which the function's value changes. It's the mathematical way to describe how a quantity varies with respect to another. Differentiation provides a powerful tool for a variety of applications, such as finding slopes of tangents, velocities, and optimizing problems.

• The derivative of a function \(f(x)\) is denoted as \(f'(x)\) or \(\frac{df}{dx}\).
• Basic rules of differentiation include the power rule, product rule, quotient rule, and chain rule.
• In our exercise, differentiating \(\cos(3-t)\) gives us \(\sin(3-t)\), which matched the initial integrand. This proves our integration solution is correct.

Checking work by differentiation is an excellent way to ensure the accuracy of your integrals.

Trigonometric Functions

Trigonometric functions are essential in calculus, particularly when dealing with integrals and derivatives. Functions like \(\sin(t)\), \(\cos(t)\), and \(\tan(t)\) are regular components in problems involving angles or periodic phenomena.

• Trigonometric identities help to simplify complex expressions in calculus problems. For example, knowing that the integral of \(\sin(u)\) is \(-\cos(u)\) and that its derivative will bring you back to \(\sin(u)\).
• Trigonometric functions have specific integral and derivative forms which can be memorized to assist in solving calculus problems faster.
• Understanding trigonometric transformations is crucial when performing the substitution and reversing processes in integration.

Incorporating trigonometric functions in calculus not only aids in solving exercises effectively but also broadens understanding of related real-world applications.