Question

Question about the fundamental theorems of calculus

Answer 1

\[d/dx \int\limits_{-3}^{1}\]

Answer 2

2t^3+3dt

Answer 3

your question doesn't seem right

and if it is you are taking derivative of a constant...

Answer 4

there are options

Answer 5

and derivative of a constant is...

Answer 6

0

Answer 7

the second equation i wrote is part of the first one

Answer 8

I don't see an equation
I see an expression

Answer 9

\[\int\limits_{-3}^{1} f(t) dt=\text{ some constant number } \\ \frac{d}{dx} \int\limits_{-3}^{1} f(t) dt=\frac{d}{dx}(\text{ some constant number} )=0\]

Answer 10

\[d/dx \int\limits_{-3}^{1} (2t ^{3}+3) dt\]

Answer 11

the choices are
a. 2t^3 + 3
b. 56.0
c. 5
d. -28.0
e. none of these

Answer 12

do you have anymore questions on this one?

Answer 13

do you understand that the definite integral is going to be a constant ?

Answer 14

by d/dx next to that means we have to differentiate that result

Answer 15

yes

Answer 16

the derivative of any constant is 0

Answer 17

yep

Answer 18

but that isnt one of the choices unless there are more steps

Answer 19

i thought you listed none of these

Answer 20

none of these means none of the above

Answer 21

and 0 wasn't listed above so it has to be none of these

Answer 22

I know that F'(x) = f(x) right?

Answer 23

yes assuming F is some antiderivative of f

Answer 24

yah

Answer 25

so would that make it option a?

Answer 26

no derivaitve of a constant is not 2t^3+3

Answer 27

or is that different because it is f(t) instead of f(x)

Answer 28

\[\int\limits_a^b f(t) dt \text{ if this exists } \\ \text{ will be a constant if the upper\lower limit is not a variable }\]

Answer 29

\[F'(x) = d/dx (\int\limits_{a}^{x} f(x) dt) = f(x)\]

Answer 30

that is one of the fundamental theorems that i think im supposed to use

Answer 31

do you understand both of your limits are constants? not one of them is a variable

Answer 32

okay so none of these

Answer 33

i would only use that if one was a variable?

Answer 34

but fine if you want to use
\[\frac{d}{dx} \int\limits_a^{x} f(t) dt=f(x) \\ \text{ then rewrite } \int\limits_{-3}^{1} f(t) dt \text{ as } \int\limits_{-3}^{x} f(t)+\int\limits_x^{1} f(t) dt =\int\limits_{-3}^{x}f(t) dt-\int\limits_{1}^{x} f(t) dt \\ \text{ so } \frac{d}{dx}(\int\limits_{-3}^{1} f(t) dt)=\frac{d}{dx}(\int\limits_{-3}^{x} f(t) dt-\int\limits_1^x f(t) dt) \\ \frac{d}{dx} \int\limits_{-3}^{1} f(t) dt=\frac{d}{dx} \int\limits_{-3}^{x } f(t) dt - \frac{d}{dx } \int\limits_1^{x} f(t) dt \\ \frac{d}{dx} \int\limits_{-3}^1 f(t) dt=f(x)-f(x) \\ \frac{d}{dx} \int\limits_{-3}^{1} f(t) dt=0\]

but all of this work is totally unnecessary given we should know that the definite integral we are differentiating is a constant

Answer 35

yah you are right thankyou

Answer 36

you could use the fundamental theorem if one or more is a variable
however if both are constant you could shove a variable in between those constants and do fundamental theorem of calculus but it is totally totally unnecessary

Answer 37

if you feel like it could you help me with one more

Answer 38

sure

Answer 39

first what is h'(x)?

Answer 40

use that fundamental theorem of calculus you were trying to mention before

Answer 41

okay one sec

Answer 42

f(t)?

Answer 43

well f(x)

Answer 44

h'(x)=f(x)

Answer 45

so we are looking at a graph of h'(x)

Answer 46

what is f(x) though?

Answer 47

f(x) is the graph which is also h'(x)

Answer 48

f(t) is the actual function so is f(x) some other function or the curve?

Answer 49

looking at the graph where is h'(3) aka f(3)

Answer 50

okay

Answer 51

-5

Answer 52

h'(3)=f(3)=-5
but choice a says h'(3)>0

Answer 53

but -5>0 is not true

Answer 54

not true okay what about the other 2?

Answer 55

one at a time

Answer 56

okay to find the critical number of h' we must look at h''
and find when h''=0


to find the critical numbers of h we must look at h'
and find when h'=0

let's do the latter first because it is much easier

Answer 57

so we are actually looking at choice 3 right now

Answer 58

h(x) has a relative maximum at x=5?
trying to see if this is true or false...

so we need to find the critical numbers of h
we start by finding h' (we done this; it is f which is the graph given)
we look to see where h'=0

Answer 59

what x values do you have h' is 0?

Answer 60

this means at what x values does h' aka f cross the x-axis

Answer 61

2,4, and 6

Answer 62

right so x=5 isn't even a contender to be a max or min since it wasn't even a critical number of h

Answer 63

wait but it is the highest point within that area so woldnt that make it a relative max?
and the global max would be at 1

Answer 64

you are looking at the graph of h' not the graph of h

Answer 65

oh okay

Answer 66

for part 2 you actually don't need to find the second derivative like I mentioned
since the graph of h' is given
the only problem is x=2 is also not a relative max of h'
h' has a global max/relative max occurring at x=1
h' has a global min/relative min occurring at x=3
h' has a relative max occurring at x=5
And there looks like h' has a relative min occurring at x=6.6 approximately

Answer 67

That first line does say the function f(t) is graphed below right?
like there isn't like a prime symbol I'm missing there right?

Answer 68

that is what it says

Answer 69

so it seems like the only one that is true is III

Answer 70

I'm pretty sure none of them are true

The critical numbers of h is when h'=0 which is when x=2,4,6

x=5 wasn't included in those numbers

Answer 71

these all seem false to me

Answer 72

oops x=0,2,4,6,8
if you want to include the endpoints

Answer 73

so none of them are true?

Answer 74

nope

Answer 75

okay well thank you for your help!!

Answer 76

weird both questions you asked are none of these

Answer 77

yah that is why i was trying to make sure that you were right because i never really get those kind of answers

Answer 78

all of these would have been right:
h'(3)<0
h'(x) has a relative maximum at x=5.
h(x) has a relative maximum at x=2.

Answer 79

wait they are all RIGHT?

Answer 80

like if the sign in the first one has been switched
or if the x=2 and x=5 has been switched in the last two options

Answer 81

no you aren't reading what I wrote if you think that :p

Answer 82

your choices were
h'(3)>0
h'(x) has a relative maximum at x=2
h(x) has a relative maximum at x=5
which are all incorrect

Answer 83

oh nevermind