Chapter 13: Problem 91
Wive an example to show that the integral of a product is not the product of the integrals.
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Chapter 13: Problem 91
Wive an example to show that the integral of a product is not the product of the integrals.
These are the key concepts you need to understand to accurately answer the question.
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A ball thrown in the air has a velocity of \(v(t)=\) \(100-32 t \mathrm{ft} / \mathrm{s}\) at time \(t\) seconds. Find the total displacement of the ball between times \(t=1\) second and \(t=7\) seconds, and interpret your answer.
Calculate the total area of the regions described. Do not count area beneath the \(x\) -axis as negative. HINT [See Example 6.] Bounded by the curve \(y=1-x^{2}\), the \(x\) -axis, and the lines \(x=-1\) and \(x=2\)
Evaluate the integrals. $$ \int_{-1.1}^{1.1} e^{x+1} d x $$
The work done in accelerating an object from velocity \(v_{0}\) to velocity \(v_{1}\) is given by $$ W=\int_{v_{0}}^{v_{1}} v \frac{d p}{d v} d v $$ where \(p\) is its momentum, given by \(p=m v(m=\) mass \()\). Assuming that \(m\) is a constant, show that $$ W=\frac{1}{2} m v_{1}^{2}-\frac{1}{2} m v_{0}^{2} $$ The quantity \(\frac{1}{2} m v^{2}\) is referred to as the kinetic energy of the object, so the work required to accelerate an object is given by its change in kinetic energy.
The marginal cost of producing the \(x\) th box of light bulbs is \(5+x^{2} / 1,000\) dollars. Determine how much is added to the total cost by a change in production from \(x=10\) to \(x=100\) boxes. HINT [See Example 5.]
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