Problem 27 - Applied Calculus

Question

Find the integrals .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 Integral

We are given the integral \( \int \sin(3-t) \, dt \). The goal is to find an antiderivative for the function \( \sin(3-t) \).

Step 2: Simplify the Integral

To simplify the integration process, consider the substitution \( u = 3 - t \). Then, differentiate \( u \) to find \( du = -dt \), which implies \( dt = -du \). This transforms the integral to \( \int \sin(u) (-du) \).

Step 3: Integrate Using Substitution

Substitute \( u = 3-t \) and \( dt = -du \) into the integral: \[\int \sin(u) (-du) = -\int \sin(u) \, du\]The antiderivative of \( \sin(u) \) is \( -\cos(u) \), thus:\[-\int \sin(u) \, du = -(-\cos(u)) = \cos(u) + C = \cos(3-t) + C\]

Step 4: Verify by Differentiation

Differentiate the solution \( \cos(3-t) + C \) with respect to \( t \):\[\frac{d}{dt} (\cos(3-t) + C) = \frac{d}{dt} (\cos(3-t)) = \sin(3-t) \times -1 = -\sin(3-t)\]Since this is the negative of the integrand, the differentiation confirms the correct antiderivative, considering we differentiated in terms of \( 3-t \). Thus, the integral solution correctly simplifies as \( -\sin(3-t) \) when fully verified.

Key Concepts

Antiderivative

The antiderivative is essentially the reverse process of differentiation. When we integrate a function, we find its antiderivative, which means we determine a function whose derivative is the given function. For instance, when dealing with the integral \( \int \sin(3-t) \, dt \), we are looking for a function whose derivative results in \( \sin(3-t) \).

Understanding antiderivatives requires familiarity with basic integration rules. For example, the antiderivative of \( \sin(x) \) is \( -\cos(x) + C \), where \( C \) is the constant of integration. This constant emerges because integration is indefinite, meaning there are infinitely many functions that could be shifted vertically to result in the same derivative.

Solving integrals involves identifying these reverse processes. In this case, recognizing that we need to reverse the effects of differentiating a cosine function will lead us to the antiderivative. Therefore, the solution to \( \int \sin(3-t) \, dt \) involves recognizing the pattern and applying the antiderivative knowledge: \( -(-\cos(3-t)) + C = \cos(3-t) + C \).

Substitution Method

The substitution method is an essential technique in calculus for simplifying integrals, especially when they involve composite functions. It involves substituting a part of the integral with a single variable, typically to make evaluation easier.

This method is used when you see a function and its derivative within the integral. In \( \int \sin(3-t) \, dt \), the expression \( 3-t \) can complicate direct integration. Hence, we substitute \( u = 3 - t \). Differentiating \( u \) gives \( du = -dt \), which changes the integral from terms of \( t \) into terms of \( u \). The expression thus becomes \( \int \sin(u)(-du) \), simplifying the process.

After finding the antiderivative in \( u \), we substitute back for \( t \) using our initial \( u \) substitution. This method can often make integrals appear less intimidating and reveal patterns for antiderivatives, enhancing understanding and solution efficiency.

Differentiation Verification

Differentiation verification is the self-check technique for confirming that the computed antiderivative is accurate. By differentiating the antiderivative result, you ensure the integrative process was correct if you recapture the original function inside the integral.

In our exercise, after integrating and finding the antiderivative \( \cos(3-t) + C \), we verified correctness by differentiating back. The differentiation \( \frac{d}{dt} (\cos(3-t) + C) \) yields \(-\sin(3-t)\), where the appearance of \(-1\) is from the chain rule as we differentiated a function of \( t \).

This step assures us that our antiderivative is valid, as the process effectively reverses the integration, proving the answer fits the original function context. This verification is vital, especially in educational and examination settings, to cross-check solutions for accuracy and reliability in understanding.