Question

Calculus problems that I got wrong on a practice help explain?

Answer 1

\[\frac{dT}{dz}=20z^3 \\ dT=20 z^3 dz \\ \text{ integrate both sides }\]

Answer 2

looks like you tried to differentiate one side

Answer 3

\[\int\limits dT =\int\limits 20 z^3 dz \\ T=\int\limits 20z^3 dz=? \\ \text{ use power rule for integrals } \\ \int\limits z^n =\frac{z^{n+1}}{n+1}+C , n \neq -1 \]

Answer 4

alright I see what I did can you help with a few more or do you want me to open another question?

Answer 5

ok why do you think that 3x^2/2+4x is not an antiderivative of 3x+4?


(3x^2/2+4x)'=6x/2+4=3x+4...

Answer 6

I didn't realize what they were talking about and now I see that it is a derivative of it.

Answer 7

\[\int\limits f(x) dx=F(x)+C \\ \int\limits (3x+4)dx=\frac{3x^2}{2}+4x+C \]

Answer 8

just pluggin the the expressions

Answer 9

but anyways lets see if we can identified the real untrue choice

Answer 10

\[f(x)=3x+4 \\ f'(x)=(3x+4)' \\ f'(x)=(3x)'+(4)' \\ f'(x)=3+0 \\ f'(x)=3 \]
first one is true

Answer 11

So for that we are just taking the derivative?

Answer 12

\[F(x)=\frac{3x^2}{2}+4x+k \\ \implies F'(x)=(\frac{3x^2}{2}+4x+k)'=\frac{6x}{2}+4+0=3x+4\]
second one is true

Answer 13

the third one...
well if f'=3
does the following seem true:
\[\int\limits 3 dx=3x+4 ?\]

Answer 14

Oh I see thanks here is another one

Answer 15

it looks like you didn't find the correct antiderivaitve of -12u or 3

Answer 16

\[\int\limits -12 u du=? \\ \int\limits 3 du=?\]

Answer 17

you could use power rule

Answer 18

\[\int\limits -12 u du =-12 \int\limits u^1 du=? \\ \int\limits 3 du=3 \int\limits u^0 du=?\]

Answer 19

6u^2+C and 3u+C

Answer 20

-6u^2+C and 3u+C
where these C's aren't necessarily the same

Answer 21

it looks like you chose blah-6u^3+3+C
instead of blah-6u^2+3u+C

Answer 22

So just explain that I found the anti derivatives incorrectly? and when i show that they are the -6u^2 include the +c?

Answer 23

On this one I just screwed up the negative and positive

Answer 24

This one I am confused.... I found the derivative and it said it was 0...

Answer 25

\[\int\limits_{a(x)}^{b(x)}g(s) ds=G(s)|_{a(x)}^{b(x)}=G(b(x))-G(a(x)) \text{ where } G'=g \\ \\ \text{ now differentiating both sides of } \\ \int\limits_{a(x)}^{b(x)} g(s) ds=G(b(x))-G(a(x)) \\ \text{ gives } \\ \frac{d}{dx} \int\limits_{a(x)}^{b(x)} g(s) ds= (G(b(x))-G(a(x)))'\]
use chain rule to differentiate the right hand side (along with difference rule)

Answer 26

54x^2 right?

Answer 27

\[\text{ should wind up with } \frac{d}{dx} \int\limits _{a(x)}^{b(x)} g(s) ds = b'(x) g(b(x))-a'(x) g(a(x)) \\ \text{ and your problem we have } \\ \frac{d}{dx} \int\limits _{-3x}^{3x} s^2 ds=(3x)'(3x)^2-(-3x)'(-3x)^2 \]

Answer 28

Alright thanks. Here is another I have this one and then one more.

Answer 29

\[3(9x^2)-(-3)(9x^2) \\ 3(9x^2)+3(9x^2) \\ 6(9x^2) \\ 54x^2 \]
yep it looks like they are showing you the correct answer each time

Answer 30

i see that I forgot to add f(0) but i dont get the /6 part

Answer 31

can you show me what you have for trapezoid rule?

Answer 32

This is what I based it off of http://www.mathwords.com/t/trapezoid_rule.htm

Answer 33

\[\Delta x =\frac{b-a}{n}\]

Answer 34

where n is the number of rectangles

Answer 35

Since it has /2 in it I should have multiplied it by 2 to get the /6?

Answer 36

sorry sub-intervals

Answer 37

we want 3 sub-intervals

Answer 38

right

Answer 39

\[\Delta x =\frac{b-a}{n}=\frac{b-a}{3}\]

Answer 40

and then there is a 2 understand delta x

Answer 41

\[\frac{\Delta x}{2} =\frac{1}{2} \Delta x =\frac{1}{2} (\frac{b-a}{3}) =\frac{b-a}{2(3)}=\frac{b-a}{6}\]

Answer 42

Ohhhhh I see it thanks. This is the last one.

Answer 43

I see that once again i forgot f(x_0)

Answer 44

and also I think the Delta x divided by a number confuses you

Answer 45

Yeah it does.

Answer 46

\[\frac{\Delta x}{c} =\frac{1}{c} \Delta x=\frac{1}{c} (\frac{b-a}{n})=\frac{1(b-a)}{c(n)}=\frac{b-a}{cn}\]

Answer 47

c here being 3

Answer 48

so after that you should have just looked at a and d
and eliminated one of those

Answer 49

also notice in simpsons rule the coefficients are symmetrical

Answer 50

Is c always 3 or is it just in this case?

Answer 51

just in the case of simpson's rule

Answer 52

can you explain that more? the part about this case being 3.

Answer 53

\[\int\limits_a^b f(x) dx=\frac{\Delta x}{3}(y_0+4y_1+2y_2+4y_3+2y_4 + \cdots + 4y_{n-1}+y_n)\]
and I know this because there is a 3 underneath the Delta x there
this is simpson's rule by the way

Answer 54

\[\frac{\Delta x}{3} =\frac{1}{3} \Delta x =\frac{1}{3} (\frac{b-a}{n})=\frac{b-a}{3n}\]
you are given that they want n to be 4

Answer 55

3n=3(4)=12

Answer 56

do you understand where the 3 is coming from on bottom?

Answer 57

\[\frac{\Delta x}{\color{red}{3}}\]

Answer 58

this factor is in simpson's rule

Answer 59

there is a 3 on bottom

Answer 60

Delta x is a fraction where it's denominator is n

Answer 61

oh I see. (I also opened a new question to work on. I tagged you. )

Answer 62

ok I will be back in a few minutes
I need coffee
it will be a few minutes

Answer 63

Alright thanks :)