asst 15

Applied Calculus II

Asst # 15

07-07-2006

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17:47:50

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**** Query problem 6.2.2 integrate x e^(-x)

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17:53:00

xe^-x - e^x + c

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17:53:02

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**** What is the indefinite integral?

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17:54:36

v du

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17:54:37

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**** What did you use for u and what for dv, what were du and v, and what integral did you obtain?

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17:55:52

u = x

du = e^-x

and

u: du=dx

v: e^x

xe^-x -e^x + c

Good, but the derivative of this expression is - x e^(-x). Be sure to check your antiderivatives by differentating them.

Details for comparison:

** We let

u = x

du = dx

dv = e^(-x)dx

v = -e^(-x)

Using u v - int(v du):

(x)(-e^(-x)) - int(-e^(-x)) dx

Integrate:

x(-e^(-x)) - (e^(-x)) + C

Factor out e^(-x):

e^(-x) (-x-1) + C. **

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17:55:53

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**** Query problem 6.2.3 integrate x^2 e^(-x)

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17:57:46

-xe^-x -2xe^-x -2e^-x +c

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17:57:47

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**** What is the indefinite integral?

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18:05:38

v du

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18:05:39

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**** For your first step, what was u and what was dv, what were du and v, and what integral did you obtain?

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18:07:12

u=x^2

dv=e^-x

du= 2dx

v=e^-x

-x^2 -2xe^-x -2e^-x +c

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18:07:12

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**** Answer the same questions for your second step.

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18:08:26

dv= e^-x dx

u=x^2

du= 2dx

v=e^-x

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18:08:26

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Applied Calculus II

Asst # 15

07-12-2006

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10:07:22

10:07:29

10:07:30

10:07:30

10:07:31

10:07:31

10:07:34

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**** Query problem 6.2.2 integrate x e^(-x) **** What is the indefinite integral? **** What did you use for u and what for dv, what were du and v, and what integral did you obtain? **** Query problem 6.2.3 integrate x^2 e^(-x) **** What is the indefinite integral? **** For your first step, what was u and what was dv, what were du and v, and what integral did you obtain? **** Answer the same questions for your second step.

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10:07:42

.starting now.

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10:07:43

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**** Query problem 6.2.18 integral of 1 / (x (ln(x))^3)

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10:13:09

ln (lnx) + C

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10:13:10

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**** What is the indefinite integral?

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10:13:25

1/u du/dx dx

** Let u = ln(x) so that du = 1 / x dx. This gives you 1 / u^3 * du and the rest is straightforward:

1/u^3 is a power function so

int(1 / u^3 du) = -1 / (2 u^2) + c.

Substituting u = ln(x) we have

-1 / (2 ln(x)^2) + c. **

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10:13:25

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**** What did you use for u and what for dv, what were du and v, and what integral did you obtain?

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10:15:12

u = 2/ x ln x dx

dv = dx

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10:15:13

10:20:22

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**** Query problem 6.2.34 (was 6.2.32) integral of ln(1+2x) **** What is the indefinite integral?

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10:24:31

i dont know.

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10:24:32

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**** What did you use for u and what for dv, what were du and v, and what integral did you obtain?

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10:24:36

i dont know.

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10:24:37

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**** Query Add comments on any surprises or insights you experienced as a result of this assignment.

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10:24:43

i dont know.

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"

See my notes and be sure you understand these procedures. Try to include more detail in the work you submit so I can see what you are thinking at each step; that would allow me to offer much better feedback.