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Chapter 7: Problem 9

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

Short Answer

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

Step by step solution

01

Identify the integral to solve

We need to evaluate the integral \( \int \sin(3-t) \, dt \). This requires finding the antiderivative of the function \( \sin(3-t) \) with respect to \( t \).
02

Use substitution method

Let \( u = 3-t \). Then, \( \frac{du}{dt} = -1 \) or \( du = -dt \). So, the integral becomes \( \int \sin(u) \, (-du) \), which simplifies to \( -\int \sin(u) \, du \).
03

Integrate \( -\sin(u) \) with respect to \( u \)

The integral of \( -\sin(u) \) is \( \cos(u) + C \), where \( C \) is the constant of integration. Thus, we have \( \cos(u) + C \).
04

Substitute back to original variable

Since \( u = 3-t \), substitute back to get \( \cos(3-t) + C \). So, the integral \( \int \sin(3-t) \, dt = \cos(3-t) + C \).
05

Differentiate to check the result

Differentiate \( \cos(3-t) + C \) with respect to \( t \). Using the chain rule, the derivative is \( - \sin(3-t) \times (-1) = \sin(3-t) \), which matches the original integrand, confirming the solution is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Antiderivative
In the world of integral calculus, finding the antiderivative is a foundational skill. The antiderivative, also known as the indefinite integral, is essentially the reverse process of differentiation. When we search for an antiderivative of a function, we want a new function whose derivative is the original function.
For example, if given the function \( \sin(3-t) \), the goal is to find another function that, when differentiated, results in \( \sin(3-t) \).
  • This involves recognizing functions that are part of standard integration rules.
  • The antiderivative includes a constant of integration represented by \( C \), highlighting that indefinite integrals have a family of solutions.
  • Identifying antiderivatives can sometimes require transformation, such as rewriting expressions into a form that matches known integrals.
By understanding and applying these concepts, we move towards solving integrals effectively.
Substitution Method
The substitution method is a crucial technique in solving integrals, especially when dealing with composite functions. This method can simplify complicated integrals by changing variables.
For instance, in our example integration problem, we have to evaluate \( \int \sin(3-t) \, dt \). By using substitution, let \( u = 3-t \), which simplifies our integral into a more manageable form.
  • First, you choose a substitution that simplifies the integrand, often setting \( u \) to an expression within the integrand.
  • Next, differentiate your chosen substitution to express \( dt \) in terms of \( du \).
  • The crucial step is to rewrite the integral with these new variables, which often reduces the problem to a basic integral with known antiderivatives.
Through substitution, we reframe the problem in a way that is more straightforward to integrate.
Chain Rule
The chain rule is a powerful tool in calculus, typically associated with differentiation. However, it shows its importance in verifying the results of an integration.
When we differentiate a composite function, the chain rule helps us find the derivative accurately. In the context of our integral, \( \cos(3-t) + C \), where \( 3-t \) is a composition with \( t \), we apply it to differentiate and verify our integration.
  • The chain rule states that \( \frac{d}{dt}[f(g(t))] = f'(g(t)) \cdot g'(t) \).
  • In our example, \( f(g(t)) \) is \( \cos(u) \) wherein \( u = 3-t \).
  • This differentiation confirms that \( \cos(3-t) \) indeed differentiates back to \( \sin(3-t) \), reaffirming our integration solution.
Understanding and applying the chain rule deepen our grasp of function behavior, both in integration and differentiation contexts.
Integration Verification
Verifying your integration is a key step in ensuring accuracy in calculus. This process involves differentiating your result and checking if it matches the original function.
In the example of \( \int \sin(3-t) \, dt \), after finding that its antiderivative is \( \cos(3-t) + C \), we differentiate \( \cos(3-t) + C \) to verify.
  • When you apply the chain rule here, you expect to reach back to \( \sin(3-t) \), the original integrand.
  • Finding that the derivative matches confirms the antiderivative is correct.
  • Verification builds confidence that the solution not only looks correct but mathematically checks out through differentiation.
This verification aligns the integrative process with differentiation, ensuring the integrity of the solution.

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Most popular questions from this chapter

In Problems \(53-58\), decide whether the statements are true or false. Give an explanation for your answer. If \(f(x)\) is a positive periodic function, then \(\int_{0}^{\infty} f(x) d x\) diverges.

(a) Between 2000 and 2010 , ACME Widgets sold widgets at a continuous rate of \(R=R_{0} e^{0.125 t}\) widgets per year, where \(t\) is time in years since January 1 2000. Suppose they were selling widgets at a rate of 1000 per year on January \(1,2000 .\) How many widgets did they sell between 2000 and \(2010 ?\) How many did they sell if the rate on January 1,2000 was 1,000,000 widgets per year? (b) In the first case ( 1000 widgets per year on January 1, 2000 ), how long did it take for half the widgets in the ten-year period to be sold? In the second case \((1,000,000 \text { widgets per year on January } 1,2000)\) when had half the widgets in the ten-year period been sold? (c) In \(2010,\) ACME advertised that half the widgets it had sold in the previous ten years were still in use. Based on your answer to part (b), how long must a widget last in order to justify this claim?

Find the solution of the initial value problem $$ y^{\prime}=\tan x+1, \quad y(0)=1 $$

(a) Find \(\int \sin \theta \cos \theta d \theta\). (b) You probably solved part (a) by making the substitution \(w=\sin \theta\) or \(w=\cos \theta .\) (If not, go back and do it that way.) Now find \(\int \sin \theta \cos \theta d \theta\) by making the other substitution. (c) There is yet another way of finding this integral which involves the trigonometric identities \(\sin (2 \theta)=2 \sin \theta \cos \theta\) \(\cos (2 \theta)=\cos ^{2} \theta-\sin ^{2} \theta\). Find \(\int \sin \theta \cos \theta d \theta\) using one of these identities and then the substitution \(w=2 \theta\). (d) You should now have three different expressions for the indefinite integral \(\int \sin \theta \cos \theta d \theta .\) Are they really different? Are they all correct? Explain.

In Problems \(51-52,\) give an example of: A function \(f(x),\) continuous at \(x=2\) and \(x=5,\) such that the integral \(\int_{2}^{5} f(x) d x\) is improper and divergent.
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