Explanation Given data: The velocity of the car (in ft/s) after time t seconds is given by v ( t ) = 6 ln ( 2 t + 1 ) . Formula used: Write the expression of position in term of velocity: s ( t ) = ∫ v ( t ) d t Here, s ( t ) is the position function of time t and v ( t ) is the velocity function of time t . Formula used for integration by parts: ∫ p d q = p q − ∫ q d p Here p and q are the functions of t. Calculation: Find the position function by using above formula. s ( t ) = ∫ v ( t ) d t = ∫ t 2 e − 2 t d t Now find ∫ t 2 e − 2 t d t by integrating it by part. Let p = t 2 and d q = e − 2 t d t . Now, d p = 2 t d t Find v by integrating d v . ∫ d q = ∫ e − 2 t d t q = − 1 2 e − 2 t Ignore the constant of integration in the above integration. Now use the above formula to find ∫ t 2 e − 2 t d t . ∫ t 2 e − 2 t d t = t 2 ( − 1 2 ) e − 2 t − ∫ ( − 1 2 ) e − 2 t ( 2 ) t d t = − 1 2 t 2 e − 2

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Chapter 13.3, Problem 41E

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To calculate: The position function from the velocity function v(t)=t2e−2t.

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Chapter 13.3, Problem 40E

Chapter 13.3, Problem 42E

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The position function from the velocity function v ( t ) = t 2 e − 2 t .

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To calculate: The position function from the velocity function v(t)=t2e−2t.

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