121616949-math.164

# 121616949-math.164 - 2 2 C Let’s suppose that at time t =...

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150 Chapter 7 Integration Exercises Find the antiderivatives of the functions: 1. 8 x 2. 3 t 2 + 1 3. 4 / x 4. 2 /z 2 5. 7 s - 1 6. (5 x + 1) 2 7. ( x - 6) 2 8. x 3 / 2 9. 2 x x 10. | 2 t - 4 | Compute the values of the integrals: 11. Z 4 1 t 2 + 3 t dt 12. Z π 0 sin t dt 13. Z 10 1 1 x dx 14. Z 5 0 e x dx 15. Z 3 0 x 3 dx 16. Z 2 1 x 5 dx 17. Find the derivative of G ( x ) = Z x 1 t 2 - 3 t dt 18. Find the derivative of G ( x ) = Z x 2 1 t 2 - 3 t dt 19. Find the derivative of G ( x ) = Z x 1 e t 2 dt 20. Find the derivative of G ( x ) = Z x 2 1 e t 2 dt Suppose an object moves so that its speed, or more properly velocity, is given by v ( t ) = - t 2 +5 t , as shown in figure 7.2 . Let’s examine the motion of this object carefully. We know that the velocity is the derivative of position, so position is given by s ( t ) = - t 3 / 3+5 t 2 / 2+ C . Let’s suppose that at time
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Unformatted text preview: 2 / 2+ C . Let’s suppose that at time t = 0 the object is at position 0, so s ( t ) =-t 3 / 3 + 5 t 2 / 2; this function is also pictured in ﬁgure 7.2 . Between t = 0 and t = 5 the velocity is positive, so the object moves away from the starting point, until it is a bit past position 20. Then the velocity becomes negative and the object moves back toward its starting point. The position of the object at t = 5 is exactly s (5) = 125 / 6, and at t = 6 it is s (6) = 18. The total distance traveled by the object is therefore 125 / 6 + (125 / 6-18) = 71 / 3 ≈ 23 . 7....
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