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Chapter 6: Problem 31

In Exercises \(22-33,\) find the general antiderivative. $$p(t)=2+\sin t$$

Short Answer

Expert verified
The general antiderivative is \(2t - \cos t + C\).

Step by step solution

01

Understand the Function

The function given is \(p(t) = 2 + \sin t\). Finding the general antiderivative means identifying a function whose derivative will give us \(p(t)\). This requires integrating each term separately.
02

Find the Antiderivative of Each Term

The integral of a constant \(2\) with respect to \(t\) is \(2t\), because the derivative of \(2t\) with respect to \(t\) is \(2\). The integral of \(\sin t\) with respect to \(t\) is \(-\cos t\), because the derivative of \(-\cos t\) is \(\sin t\).
03

Combine the Antiderivatives

Combine the results from Step 2 to write the general antiderivative: \[\int (2 + \sin t) \, dt = 2t - \cos t + C\]Here, \(C\) is the constant of integration, representing any constant term that could be added to an antiderivative.

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Key Concepts

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

Integral Calculus
Integral calculus is an essential branch of calculus that deals with finding antiderivatives or integrals. When we take the integral of a function, we're essentially finding out which function, when differentiated, will give us the original integrand. This is often used to compute things like area under curves and solve differential equations. In our specific exercise, we apply integral calculus to the function \(p(t) = 2 + \sin t\), breaking it down into simpler parts to integrate each term separately.

Here's a simple way to understand it:
  • Identify each term in the function you want to integrate.
  • Use known integral rules to find the antiderivative of each part.
  • Combine these results to get the total antiderivative.
By mastering these basic steps, you can apply integral calculus to a wide range of functions.
Constant of Integration
When finding an antiderivative, we often end up with not just a single result, but a family of functions. This family is defined by adding an arbitrary constant, represented by \(C\), called the constant of integration. Since the derivative of any constant is zero, we have this unknown constant in indefinite integrals.

Here's why we need it:
  • Differentiating a constant gives zero.
  • Any integral thus could have had any constant added to it initially.
  • The integration process brings back these unknown constants as \(C\).
So in our exercise, when we integrate \(2 + \sin t\), we add \(C\) to the result, obtaining \(2t - \cos t + C\). This represents all possible antiderivatives of the given function.
Trigonometric Integrals
Trigonometric integrals involve integrating functions that include trigonometric functions such as \(\sin x, \cos x\), and more. Understanding how to work with these is crucial in calculus since these functions often model various natural phenomena.

For instance, consider the integration of \(\sin t\):
  • The integral of \(\sin t\) is \(-\cos t\).
  • We add the constant of integration \(C\) after integrating.
  • Knowing the derivatives of basic trigonometric functions helps in quickly identifying their integrals.
Mastering trigonometric integrals equips you to tackle more complex integrals and apply calculus to solve real-world problems that involve periodic functions, such as oscillations and waves.

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Most popular questions from this chapter

In Exercises \(22-33,\) find the general antiderivative. $$h(x)=4 x^{3}-7$$

For Problems \(97-99,\) let \(\int g(x) d x=G(x)+C .\) Which of (I)-(III), if any, is equal to the given integral? \(\int \cos (G(x)) g(x) d x\) I. \(\sin (G(x)) g(x)+C\) II. \(\sin (G(x)) G(x)+C\) III. \(\sin (G(x))+C\)

In Exercises \(6-21,\) find an antiderivative. $$f(t)=2 t^{2}+3 t^{3}+4 t^{4}$$

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Are the statements in Problems \(104-112\) true or false? Give an explanation for your answer. If \(F(x)\) and \(G(x)\) are two antiderivatives of \(f(x)\) for \(-\infty < x < \infty\) and \(F(5) > G(5),\) then \(F(10) > G(10)\).
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