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Mth252 Test#2 Review Ch4.9, 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, 5.7. 1 Name__________________ Ch4.9 Antiderivatives 1. If g ''( x ) = 5cos x () , g(0) = 1, and g '(0) = 0, find g p () . 2. A particle moves with acceleration function 2 2 4 5 ) ( t t t a - + = . Its initial velocity is v(0) = 3 m/s and its initial displacement is s(0)=10 m. Find its position after t seconds. Ch5.1 Areas and Distance 3. When we estimate distances from velocity data it is sometimes necessary to use times to, t 1 , t 2 , t 3 , ...that are not equally spaced. We can still estimate distances using the time periods Dt i = t i –t i=1 . For example, on May 7, 1992, the space shuttle Endeavor was launched on mission STS-49, the purpose of which was to install a new perigee kick motor in an lntelsat communications satellite. The table, provided by NASA, gives the velocity data for the shuttle between liftoff and the jettisoning of the solid rocket boosters. Event Time (s) Velocity (ft/s) Launch 0 0 Begin roll maneuver 10 185 End roll maneuver 15 319 Throttle to 89% 20 447 Throttle to 67% 32 742 Throttle to 104% 59 1325 Maximum dynamic pressure 62 1445 Solid rocket booster separation 125 4151 Use these data points to estimate the height above Earth’s surface of the space shuttle Endeavor, 62 seconds after liftoff. Calculate using left endpoints and right endpoints. Make a sketch. Ch5.2 p. The Definite Integral 4. Use the graph in figure 1 to evaluate each integral. Show work. a) Ú 2 0 ) ( dx x g b) Ú 6 20 ) ( dx x g c) Ú 7 0 ) ( dx x g figure 1. 5. Suppose 4 dx and 6 8 5 3 3 0 2 0 = = = Ú Ú Ú f(x) , dx ) x ( f , dx ) x ( f . Find the following: (a) f x dx () 0 5 Ú (b) f x dx () 2 3 Ú (c) f x dx () 5 2 Ú

Mth252 Test#2 Review Ch4.9, 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, 5.7.spot.pcc.edu/~jkissick/summer/math252exam2review.pdf · 5 3 3 0 2 0 Úf(x)dx=,Úf(x)dx=, ... Find the following: (a)

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Mth252 Test#2 Review Ch4.9, 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, 5.7.

1

Name__________________Ch4.9 Antiderivatives1. If

g ' '(x) = 5cos x( ), g(0) =1, and g '(0) = 0, find g p( ).

2. A particle moves with acceleration function 2245)( ttta -+= . Its initial velocity is v(0) = 3 m/s and its initialdisplacement is s(0)=10 m. Find its position after t seconds.

Ch5.1 Areas and Distance3. When we estimate distances from velocity data it is sometimes necessary to use times to, t1, t2, t3, ...that are notequally spaced. We can still estimate distances using the time periods Dti = ti –ti=1. For example, on May 7, 1992, thespace shuttle Endeavor was launched on mission STS-49, the purpose of which was to install a new perigee kick motor inan lntelsat communications satellite. The table, provided by NASA, gives the velocity data for the shuttle between liftoffand the jettisoning of the solid rocket boosters.

Event Time (s) Velocity (ft/s)Launch 0 0Begin roll maneuver 10 185End roll maneuver 15 319Throttle to 89% 20 447Throttle to 67% 32 742Throttle to 104% 59 1325Maximum dynamic pressure 62 1445Solid rocket booster separation 125 4151

Use these data points to estimate the height above Earth’s surface of the space shuttle Endeavor, 62 seconds after liftoff.Calculate using left endpoints and right endpoints. Make a sketch.

Ch5.2 p. The Definite Integral4. Use the graph in figure 1 to evaluate each integral. Show work.

a) Ú2

0)( dxxg b) Ú

6

20)( dxxg c) Ú

7

0)( dxxg

figure 1.

5. Suppose 4dx and 6 85

3

3

0

2

0

=== ÚÚÚ f(x),dx)x(f,dx)x(f . Find the following:

(a) f x dx( )0

5

Ú (b) f x dx( )2

3

Ú (c) f x dx( )5

2

Ú

Mth252 Test#2 Review Ch4.9, 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, 5.7.

2

6. a) Use ÂÚ=

*

•ÆD=

n

ii

n

b

axxfdxxf

1

)(lim)( to evaluate the integral Ú +2

0)1( dxx

b) Verify using the evaluation theorem.c) Illustrate graphically with area under a curve.

Ch5.3 Evaluating Definite Integrals7. Let r(t) be the rate at which the world’s oil is consumed, where t is measured in years starting at t=0 on January 1,

2000, and r(t) is measured in barrels per year. Write a sentence to explain what Ú3

0)( dttr means?

Ch5.4 The Fundamental Theorem of Calculus

8. Let g x f t dtx

( ) ( )=-Ú3

, 33 ££- x , where f is the function whose graph is shown in Figure 2.

Figure 2a) Evaluate )3( g(1), ),0(),1( ),3( gggg -- .b) On what interval(s) is g increasing? On what interval(s) is g decreasing?c) Where does g attain its absolute maximum value?d) On what intervals is g concave downward?e) Sketch the graph of g.

Ch5.5 The Substitution Rule

9. Úp 2/

0

)sin()cos( dxex x

Ch5.6 Integration by Parts10 dxxxÚ )ln(5

Ch 5.7 Additional Techniques of Integration

11. Ú ⋅ dxxx )(cos)(sin 43 12.

5x - 314x 2 +11x -15

Ê

Ë Á

ˆ

¯ ˜

1

2

Ú dx What happens if the limits of integration

are 0 to 1?

13.

x 3 9 - x 2( )0

3

Ú dx Hint : Problem can be worked with 2 different methods

Method 1: x = 3sin q( ), followed by u = cos q( )Method 2 : u = 9 - x 2, and solve for x 2

y f t= ( )