Use basic integration formulas to compute the following antiderivatives. \(\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right) d x\) Text Transcription: int (sqrt x-1/sqrt x) dx
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Textbook Solutions for Calculus Volume 1
Question
A horizontal cylindrical tank has cross-sectional area \(A(x)=4\left(6 x-x^{2}\right) m^{2}\) at height x meters above the bottom when \(x \leq 3\).
(a) The volume V between heights a and b is \(\int_{a}^{b} A(x) d x\). Find the volume at heights between 2 m and 3 m.
(b) Suppose that oil is being pumped into the tank at a rate of 50 L/min. Using the chain rule, \(\frac{d x}{d t}=\frac{d x}{d V} \frac{d V}{d t}\), at how many meters per minute is the height of oil in the tank changing, expressed in terms of x, when the height is at x meters?
(c) How long does it take to fill the tank to 3 m starting from a fill level of 2 m ?
Text Transcription:
A(x)=4(6x-x^2) m^2
x leq 3
int_a^b A(x) dx
dx/dt=dx/d V d V/dt
Solution
The first step in solving 5.4 problem number trying to solve the problem we have to refer to the textbook question: A horizontal cylindrical tank has cross-sectional area \(A(x)=4\left(6 x-x^{2}\right) m^{2}\) at height x meters above the bottom when \(x \leq 3\).(a) The volume V between heights a and b is \(\int_{a}^{b} A(x) d x\). Find the volume at heights between 2 m and 3 m.(b) Suppose that oil is being pumped into the tank at a rate of 50 L/min. Using the chain rule, \(\frac{d x}{d t}=\frac{d x}{d V} \frac{d V}{d t}\), at how many meters per minute is the height of oil in the tank changing, expressed in terms of x, when the height is at x meters?(c) How long does it take to fill the tank to 3 m starting from a fill level of 2 m ?Text Transcription:A(x)=4(6x-x^2) m^2x leq 3int_a^b A(x) dxdx/dt=dx/d V d V/dt
From the textbook chapter Integration Formulas and the Net Change Theorem you will find a few key concepts needed to solve this.
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