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Chapter 8: Problem 8

\(h(t)=\int_{0}^{t} \sin v d v\)

Short Answer

Expert verified
h(t) = 1 - \cos t

Step by step solution

01

Understand the Problem

The integral represents the function of the variable t which is given by the integral of sin(v) with respect to v from 0 to t.
02

Set up the Integral

Identify that we need to integrate the function \( \sin v \) with respect to v over the interval from 0 to t. The integral is written as \[ h(t) = \int_{0}^{t} \sin v \ d v \].
03

Integrate the Function

Integrate \( \sin v \). The antiderivative of \( \sin v \) is \(- \cos v \).
04

Apply the Limits of Integration

Evaluate the integral at the bounds 0 and t: \[ h(t) = \left[ - \cos v \right]_{0}^{t} = - \cos t - (- \cos 0). \]
05

Simplify the Expression

Simplify the expression: \[ h(t) = - \cos t + \cos 0. \] Since \(\cos 0 =1 \), the final answer is: \[ h(t) = 1 - \cos t \].

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Key Concepts

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

Definite Integrals
In integral calculus, a definite integral has specific bounds. It represents the accumulation of quantities, like the total area under a curve, between two points. Unlike an indefinite integral, which lacks limits, a definite integral offers a precise numerical value once evaluated. Use integrals to find sums, areas, and other total measures. Consider the integral \(\int_{0}^{t} \sin v \ dv\). Here, 0 and t are the bounds. This tells us to accumulate the area under \(\textrm{sin} \ v\) from v=0 to v=t.
Trigonometric Functions
Trigonometric functions like \(\textrm{sin}, \cos\) and others, often appear in calculus problems. These functions describe relationships in triangles and have unique wave patterns. They repeat every 2蟺 (periodicity). In the example \(\textrm{sin} \ v\), the sine function oscillates between -1 and 1. Its graph starts at 0, peaks at 1 (蟺/2), goes back to 0 (蟺), reaches -1 (3蟺/2), and returns to 0 (2蟺). Integration looks at the area under these waves.
Antiderivatives
An antiderivative reverses differentiation, representing a function whose derivative is the original function. If you know \(\textrm{f}(x)\) and its derivative is \(\textrm{f}'(x)\), then \(\textrm{f}(x)\) is the antiderivative. For example, the derivative of \(\textrm{sin} \ x\) is \(\textrm{cos} \ x\); thus, by integrating \(\textrm{cos} \ x\), we get the antiderivative \(\textrm{sin} \ x\). In our exercise, the integral of \(\textrm{sin} \ v\) is \(- \cos v\). This means differentiating \(- \cos v\) gives \(\textrm{sin} \ v\). So, integrating helps track back to the original function.

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Most popular questions from this chapter

\(x^{\prime \prime}+y^{\prime \prime}=x+t, y^{\prime}-x+x^{\prime}=0, x(0)=x^{\prime}(0)=\) \(y(0)=y^{\prime}(0)=0\)

\(x^{\prime}=x+y+f(t), y^{\prime}=-2 x-2 y, x(0)=\) \(y(0)=0\), where \(f(t)=\left\\{\begin{array}{l}1,0 \leq t<1 \\ -1,1 \leq t<2\end{array}\right.\) and \(f(t)=f(t-2)\) if \(t \geq 2\)

\(x^{\prime}+7 x+4 y=f(t), y^{\prime}+6 x-3 y=0, x(0)=\) \(y(0)=0\), where \(f(t)=\left\\{\begin{array}{l}1,0 \leq t<1 \\ 0, t \geq 1\end{array}\right.\)

\(h(t)=\int_{0}^{t} v \cos (t-v) d v\)

\(y^{\prime \prime}+3 t y^{\prime}-6 y=0, y(0)=1, y^{\prime}(0)=0\)
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