Question

PLEASE HELP I'M COMPLETELY DESPERATE...
I have to finish this before 8AM in order to graduate but I'm so confused and frustrated with myself, please help I'm begging.
- http://prntscr.com/bcx16a
- http://prntscr.com/bcx1od

Answer 1

@Michele_Laino

Answer 2

for the second question, the partitions have width 1.
If the function is decreasing and each partition is evaluated on the right
(i.e. the upper right corner intercepts the curve)
or if the function is increasing and each partition is evaluated on the left,
(i.e. the upper left corner intercepts the curve)
then you will underestimate the integral

Answer 3

hint:
please integrate this:
\[\int {\frac{{dy}}{{\sqrt {1 - {y^2}} }}} = \int {6xdx} \]

Answer 4

wat? how does that help?

Answer 5

For the first problem? Okay one sec

Answer 6

hint:
try this substitution:
\(y=\sin\theta\)
wherein \(\theta\) is the new variable

Answer 7

oh the first question

Answer 8

yeah it is a separable differential equation

Answer 9

wait is it just 3x^2?

Answer 10

ohhhh

Answer 11

no

Answer 12

you have to add the arbitrary constant at the right side @tmagloire1

Answer 13

i'm confused not sure which side to integrate

Answer 14

\[\int\frac{dy}{\sqrt{1-y^2}}=\int6x\;dx
\\\int\frac{cos\;\theta\;d\theta}{cos\;\theta}=3x^2
\\\theta=3x^2
\\y=sin^{-1}(3x^2)\]

Answer 15

hint:
please you have to integrate both sides
right side = \(\int {6xdx} = 3{x^2} + C,\quad C \in \mathbb{R}\)

Answer 16

that's already separable?

Answer 17

so you were close, but forgot to back substitute

Answer 18

so y = inverse sin (3x^2) is the solution for part A or is that just part of it?

Answer 19

if we integrate both sides, we get:
\[\arcsin y = 3{x^2} + C,\quad C \in \mathbb{R}\]

Answer 20

yeah don't forget the arbitrary constant. and that is part A

Answer 21

for part B, what are the domains of x and y?

Answer 22

Part B just requires an explanation they don't give us that information just the equation and y(0)

Answer 23

is it because they're inseparable possibly?

Answer 24

hint:
we can rewrite the solution, as follows:
\[y = \sin \left( {3{x^2} + C} \right),\quad C \in \mathbb{R}\]

Answer 25

well, let's start by answering the questions "What are the allowable values of x?" and "What are the allowable values of y?"

Answer 26

now, if we apply the initial condition, we get:
\[2 = \sin C,\quad C \in \mathbb{R}\]
is that condition possible?

Answer 27

for y it's 2 and for x we don't know x

Answer 28

no it isn't?

Answer 29

why?

Answer 30

no that is not the way to approach part B

Answer 31

You must prove that y(0)=2 is not in the domain of the curve

Answer 32

How do I do that?

Answer 33

again, I will ask "what are the possible values of x?"

Answer 34

not sure how to figurer that out

Answer 35

ok, are there any values of x for which \(6x\sqrt{1-y^2}\) is undefined?

Answer 36

0 and 2?

Answer 37

no, all real x are ok (i.e. there is no restriction on x)

Answer 38

what about y?

Answer 39

2 for y

Answer 40

on what interval is \(\sqrt{1-y^2}\) undefined?

Answer 41

6x is defined for all x

Answer 42

1 and -1

Answer 43

no, those values are ok. where is a square root defined for real numbers?

Answer 44

the negative or positive of the same number?

Answer 45

im gettitng confused

Answer 46

no it is defined when the argument (thing under the radical) is non-negative

Answer 47

So how does that prove that the argument doesn't have a solution if it's defined?

Answer 48

is it defined at y=2? is \(\sqrt{1-2^2}\) a real number?

Answer 49

Oh no it isn't

Answer 50

exactly

Answer 51

So it doesn't have a solution because the argument is undefined

Answer 52

right

Answer 53

Ohh gotcha, thank you! I just need help with the last trapezoid problem now

Answer 54

ok. basically a trapezoidal region averages the lower bound and upper bound of the area of each partition

Answer 55

ok