Question

Consider the function f(x) = 6sin(x2) on the interval 0 ≤ x ≤ 3.

(a) Find the exact value in the given interval where an antiderivative, F, reaches its maximum.
x = sqrt(pi)

(b) If F(1) = 5, estimate the maximum value attained by F. (Round your answer to three decimal places.)
… I just need help with part b; I don'y know how to begin

Answer 1

its f(x)=6sin(x^2) not "x2"… sorry :/

Answer 2

so the first question sounds like you want to find
\[\frac{d}{dx}\int\limits_{0}^{x}f(t) dt=0\]
this will help you find the critical numbers of F

Answer 3

find when x satisfies the equation:*

Answer 4

basically solve 6sin(x^2)=0 for x

Answer 5

i did i got sort (pi) for a

Answer 6

sqrt*

Answer 7

I guess you already did a
I got 2 zeros to 6sin(x^2)=0 on [0,3]
then you can plug them into the derivative of 6sin(x^2) to see if it is a max or not

Answer 8

so it'd be -6cos(x^2)*2x = -12cos(x^2)? I just plug the two 0's into that and see what the max value is right..?

Answer 9

where you get the negative at?

Answer 10

\[12xcos(x^2)\]

Answer 11

oh my bad its just cos, i got confused with the -cos

Answer 12

i was thinking about antiderivatives

Answer 13

yep one of the zeros gives a min and one of them gives a max

Answer 14

and you chose the right zero that will give us a max
gj

Answer 15

for 0 i got 0 and for 3 i got -32.80068943

Answer 16

and for sqrt(pi) i got -21.26944621

Answer 17

\[6 \sin(x^2)=0 \\ \sin(x^2)=0 \\ x^2=n \pi\]
where n is integer greater than 0
we want x to be between 0 and 3
so we want x^2 to be between 0 and 9
so n could be 1 or 2
xo we x=sqrt(pi) or x=sqrt(2pi)

Answer 18

but 12sqrt(pi)cos(pi) is negative so at x=sqrt(pi) we have a max
and 12sqrt(2pi)cos(2pi) is positive so at x=sqrt(2pi) we have a min

Answer 19

so it want's us to use F(1)=5 to estimate F(sqrt(pi))

Answer 20

thats where i don't know what to do:/

Answer 21

\[F(x)=\int\limits_{0}^{x}6 \sin(t^2) dt \]
we have
\[F(1)=\int\limits_{0}^{1}6 \sin(t^2) dt =5 \\ F(\sqrt{\pi})=\int\limits_{0}^{\sqrt{\pi}}6 \sin(t^2) dt=n\]
n being the value we want to obtain...

Answer 22

so finding the area under the curve of f(x) using the interval given will give me the value of n..?

Answer 23

we have have (1,5) is on F
we can find the slope at (1,5) of F

Answer 24

\[y-y_1=m(x-x_1) \\ y=m(x-x_1)+y_1 \]
my idea is to maybe a build a tangent line to F at (1,5)

Answer 25

hmm okay

Answer 26

you already found the derivative of F
use that to find the slope at x=1

Answer 27

6.48362767

Answer 28

is that the slope?

Answer 29

yes

Answer 30

I got 6sin(1) is approximately 5.0488

Answer 31

oh because i plugged in the value into the derivative of the function, not the original function

Answer 32

\[F(x) \approx 5.0488(x-1)+5 \\ \text{ this is usually a pretty good approximation for values near x=1 } \\ \]

Answer 33

well F'=f right?

Answer 34

the derivative of big F is small f

Answer 35

true!!!

Answer 36

I'm just worried about how good that approximation will be since sqrt(pi) is a little further than 1 than I would like

Answer 37

but I can't see how to get a better approximation right now

Answer 38

i see what you're saying, and exactly what i was thinking...

Answer 39

however if we do use that "approximation" we will get 8.89997

Answer 40

Have you learned any other approximation techniques?

Answer 41

i am afraid not

Answer 42

sounds like the linear approximation is the way to go then

Answer 43

okay i will try that!

Answer 44

is this an online assignment?

Answer 45

well i think it said round to 3 decimal places

Answer 46

yes it was wrong :/ its a tough question

Answer 47

https://answers.yahoo.com/question/index?qid=20101120210256AAl0I6z
this person did this...

Answer 48

I don't know what technique that is

Answer 49

they had F(1)=7

Answer 50

one of my classmates had that and they inputed that answer and it was wrong as well

Answer 51

did you enter in 9?

Answer 52

I was going to try that next haha

Answer 53

because 8.89997
to 3 decimal poitns shuold be 9.000 or just 9

Answer 54

9 was wrong as well

Answer 55

you get infinite amount of answers?

Answer 56

hmm i think I'm going to move onto another question… will you be willing to help me with another question..? Im a bit lost in this chapter because i wasn't able to attend the lecture :/

Answer 57

lol try this
1.169+5

Answer 58

6.169

Answer 59

haha okay i'll give it a shot lol

Answer 60

lol it was wrong too

Answer 61

ill post the next question

Answer 62

I'm sorry I put in the wrong integral into wolfram I put 2sinx^2

Answer 63

\[\int\limits\limits_{1}^{\ \sqrt{\pi}}6 \sin(x^2)+F(1) \\ 3.507+5=8.507\]

Answer 64

not too far off from our approximation