section 5.5 hw

Current Score : 37 / 39 Due : Thursday, April 24 2014 11:59 PM EDT

1. 1/1 points | Previous Answers SCalcET7 5.5.001.

Evaluate the integral by making the given substitution. (Use C for the constant of integration.)

Section 5.5 HW (Homework)Frances CoronelMAT 151 Calculus I, Spring 2014, section 01, Spring 2014Instructor: Ira Walker

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e−11x dx, u = −11x

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2. 1/1 points | Previous Answers SCalcET7 5.5.002.MI.


Master ItEvaluate the integral by making the given substitution.

Part 1 of 4We know that if then

Therefore, if then

x3(9 + x4)4 dx, u = 9 + x4

x3(2 + x4)5 dx, u = 2 + x4

u = f(x), du = f '(x) dx.

u = 2 + x4, du = (No Response) dx.





Part 1 of 4

We know that if then Therefore, if then

4. 4/4 points | Previous Answers SCalcET7 5.5.003.MI.SA.

This question has several parts that must be completed sequentially. If you skip a part of the question,you will not receive any points for the skipped part, and you will not be able to come back to the skippedpart.

Tutorial ExerciseEvaluate the integral by making the given substitution.

Step 1

We know that if then Therefore, if then

Step 2

If is substituted into then we have

We must also convert into an expression involving u.

x2 dx, u = x3 + 33x3 + 33

x2 dx, u = x3 + 20x3 + 20

u = f(x), du = f '(x) dx. u = x3 + 20,du = (No Response) dx.

x2 dx, u = x3 + 18x3 + 18

u = f(x), du = f '(x) dx. u = x3 + 18,

du = dx.

u = x3 + 18 ,x2 dxx3 + 18 = .x2 (u)1/2 dx u1/2 x2 dx

x2 dx


We know that and so

Step 3

Now, if then

This evaluates as

Step 4

Since then converting back to an expression in x we get

You have now completed the Master It.

du = 3x2 dx, x2 dx = du.

u = x3 + 18, = = .x2 dxx3 + 18 u1/2 du13

13 u1/2 du

= + C.13 u1/2 du

u = x3 + 18,

u3/2 + C = .29


5. 1/1 points | Previous Answers SCalcET7 5.5.004.





Part 1 of 4We know that if then

Therefore, if then

7. 4/4 points | Previous Answers scalcet7 5.5.006.mi.sa.nva


, u = 1 − 6tdt(1 − 6t)9

dx, u = 1/x7sec2(1/x7)x8

dx, u = 1/x2sec2(1/x2)x3

u = f(x), du = f '(x) dx.

u = ,1x2 du = (No Response) dx.



Tutorial ExerciseEvaluate the integral by making the given substitution.

Step 1We know that if then

Therefore, if then

Step 2



We know that and so

Step 3

Now, if then

This evaluates as

Step 4



dx, u = 1/x2sec2(1/x2)x3

u = f(x), du = f '(x) dx.

u = ,1x2 du = dx.

u = 1x2 , dxsec2 (1/x2)

x3 = . dxsec2 u

x3sec2 u dx1

x3

dx1x3

du = dx,−2x3 dx = du.1

x3

u = ,1x2 = = − . dxsec2(1/x2)

x3

sec2 u − du12

12 sec2 u du

− = + C.12 sec2 u du

u = ,1x2

− tan u + C = 12


Evaluate the indefinite integral. (Use C for the constant of integration.)

Master ItEvaluate the indefinite integral.

Part 1 of 4We must decide what to choose for u.

If then and so it is helpful to look for some expression in

for which the derivative is also present, though perhaps missing a constant factor.

For example, is part of this integral, and the derivative of is (No Response) , which is alsopresent except for a constant.

9. 4/4 points | Previous Answers scalcet7 5.5.007.mi.sa.nva


Tutorial ExerciseEvaluate the indefinite integral.

Step 1

x17 sin(x18) dx

x12 sin(x13) dx

u = f(x), du = f '(x) dx, x12 sin(x13) dx

x13 x13

x4 sin(x5) dx


We must decide what to choose for u.

If then and so it is helpful to look for some expression in for

which the derivative is also present, though perhaps missing a constant factor.

For example, is part of this integral, and the derivative of is , which is alsopresent except for a constant.

Step 2

If we choose then



We know that and so

Step 3

Now, if then

This evaluates as

Step 4



u = f(x), du = f '(x) dx, x4 sin(x5) dx

x5 x5

u = x5, du = 5x4 dx.

u = x5 ,x4 sin(x5) dx = .x4 sin u dx sin u(x4 dx)

x4 dx

du = 5x4 dx, x4 dx = du.

u = x5, = = .x4 sin(x5) dx sin u du15

15

sin u du

= + C.15

sin u du 15

u = x5,

− cos u + C = .15

10.0/1 points | Previous Answers SCalcET7 5.5.010.


11.1/1 points | Previous Answers SCalcET7 5.5.013.MI.

Evaluate the indefinite integral. (Remember to use ln |u| where appropriate. Use C for the constant ofintegration.)




for which the derivative is also present, though perhaps missing a constant

factor.

We see that is part of this integral, and the derivative of is (No Response) , which issimply a constant.

(8t + 1)2.9 dt

dx7 − 5x

dx8 − 6x

u = f(x), du = f '(x) dx,

= dx8 − 6x

dx18 − 6x

8 − 6x 8 − 6x



12.4/4 points | Previous Answers SCalcET7 5.5.013.MI.SA.



Step 1We must decide what to choose for u.


for which the derivative is also present, though perhaps missing a constant

factor.

We see that is part of this integral, and the derivative of is -7 , which is simply aconstant.

Step 2If we choose then



Using then we get

Step 3

Now, if then

This evaluates as the following. (Remember to use ln |u| where appropriate.)

dx4 − 7x

u = f(x), du = f '(x) dx,

= dx4 − 7x

dx14 − 7x

4 − 7x 4 − 7x

u = 4 − 7x, du = −7 dx.

u = 4 − 7x ,dx4 − 7x

= .dx4 − 7x

dx1u

dx

du = −7 dx, dx = du.

u = 4 − 7x, = = − .dx4 − 7x

− du1u

17

17

du1u

− = + C17

du1u

Step 4Since then converting back to an expression in x we get the following. (Remember to useln |u| where appropriate.)






u = 4 − 7x,

− ln|u| + C = 17

sin 5πt dt

dueu

(3 − eu)2











for which the derivative is also present, though perhaps missing a

constant factor.

For example, is part of this integral, and the derivative of is (No Response) , which is alsopresent.

dxa + bx20

21ax + bx21

dx(ln x)24

x

dx(ln x)21

x

u = f(x), du = f '(x) dx,

= dx(ln x)21

x(ln x)21 dx1

x

ln x ln x








for which the derivative is also present, though perhaps missing a

constant factor.

For example, is part of this integral, and the derivative of is , which isalso present.

Step 2

If we choose then


We must also convert into an expression involving u, but we know already that

Step 3

Now, if then

dx(ln x)41

x

u = f(x), du = f '(x) dx,

= dx(ln x)41

x(ln x)41 dx1

x

ln x ln x

u = ln x, du = dx.1x

u = ln x , dx(ln x)41

x

= . dx(ln x)41

xu41 dx1

x

dx1x

dx = du.1x

u = ln x, = dx(ln x)41

xu41 du.

This evaluates as

Step 4Since then converting back to an expression in x we get




= + C.u41 du

u = ln x,

u42 + C = .142

sec2 θ tan3 θ dθ










for which the derivative is also present.

We see that is part of this integral, and the derivative of is (No Response) , which isalso present.

sin(5 + x7/2) dxx5

ex dx21 + ex

ex dx22 + ex

u = f(x), du = f '(x) dx, ex dx22 + ex

22 + ex 22 + ex








for which the derivative is also present.

We see that is part of this integral, and the derivative of is ,which is also present.

Step 2

If we choose then


We must also convert into an expression involving u, but we already know that

Step 3

Now, if then

ex dx48 + ex

u = f(x), du = f '(x) dx, ex dx48 + ex

48 + ex 48 + ex

u = 48 + ex, du = ex dx.

u = 48 + ex ,ex dx48 + ex

= = ex dx48 + ex ex dxu (ex dx)u

ex dx

ex dx = du.

u = 48 + ex, = = .ex dx48 + ex duu u1/2 du

This evaluates as

Step 4



= + C.u1/2 du

u = 48 + ex,

u3/2 + C = .23


section 5.5 hw

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