Question

Find the area of the region between y=x^(1/2) and y=x^(1/3) for 0≤x≤1

Answer 1

can you draw them out?

Answer 2

go to here and type in the functions https://www.desmos.com/calculator

Answer 3

Im interested in understanding the concepts behind the problem- I have the answer (.0833)
This is in calculus, using definite integrals.

Answer 4

No graph is included in the question.

Answer 5

So the formula is \[\int\limits_{a}^{b} TOP function - BOTTOM function\]

Answer 6

@VostheBoss321 but we have to know which one is upper function in interval.

Answer 7

f(b)-f(a)?

Answer 8

You can graph it yourself and you probably know what x^1/2 looks like

Answer 9

I drew it, take a look, and follow ConDawg https://www.desmos.com/calculator

Answer 10

I know what both graphs look like, but is the only way to sole this problem graphically? Or is there an analytical method?

Answer 11

You need to graph it first it help to understand things a lot better and it doesnt take long.

Answer 12

you see,in your interval, 0,1, upper is x^1/3 , out of the interval, upper is x^1/2. No way to solve without graph

Answer 13

go to that website and type in
y=x^1/2
y=x^1/3
it takes two seconds

Answer 14

Alright, I'm looking at the graph from site Condawg provided, and I see the space in question

Answer 15

How do I compute the area of the region though?

Answer 16

\[\int\limits_{0}^{1}x ^{1/3}-x ^{1/2}\]

Answer 17

That simple?
Just integral f(b)-f(a)?

Answer 18

integral of the top function (x^1/3) minus the integral of the bottom function (x^1/2)

Answer 19

Piece of cake, eh?

Answer 20

Yup! Thank you both very much!

Answer 21

tell me what you get for an answer though!

Answer 22

What value should I be using for x in this case? Or do I apply power rule? (1 and 0 raised to any power end up canceling each other out in above expression)

Answer 23

Do you know how to take the integral of x^1/2 and x^1/3?

Answer 24

Frankly, ive been at this a few hours now and I feel theyve slipped my mind

Answer 25

I do understand how to take integrals though

Answer 26

Don't worry I'll show you!

Answer 27

in this case, is it 1/1.5x^1.5 and 1/1.3x^1.3?

Answer 28

+c of course

Answer 29

This is the first step is to write it like this! Write this down on your paper!
\[\int\limits_{0}^{1}x ^{1/3}-\int\limits_{0}^{1}x ^{1/2}\]

Answer 30

P.S. use don't use "c" for definite integrals, but that's irrelevant.

Answer 31

I understand, thats implicit in the whole definite integral concept, correct?

Answer 32

Yeah, but I'm going to go through this step by step, and you write each step down. So write step one down.

Answer 33

Done

Answer 34

Step two: evaluate the integrals
To evaluate these integrals we are going to use a method I call the anti derivative method.

This method is add 1 to the exponent and divide the whole thing by the new exponent.
So: \[\frac{ 3 }{ 4 }*x ^{\frac{ 4 }{ 3 }}0 \to 1\] and \[\frac{ 2 }{ 3 }x ^{\frac{ 3 }{ 2 }} 0 \to 1\]

Answer 35

Questions?

Answer 36

I need to look at the anti-derivative method again, we have learned it but in a later chapter.

Answer 37

But I understand your logic herein completely.

Answer 38

Step 3: evaluate both the integrals from 0 to 1 then subtract the top function by the bottom function

to evaluate the 3/4*x^4/3 we plug in 1 then evaluate then subtract that by what we get by plugging 0 and evaluating.

So: \[([\frac{ 3 }{ 4 }(1)^{\frac{ 4 }{ 3 }}) - (\frac{ 3 }{ 4 }(0)^{\frac{ 4 }{ 3 }})]\] that's for the top function

the bottom function: \[\frac{ 2 }{ 3 }(1)^{\frac{ 3 }{ 2 }} - \frac{ 2 }{ 3 }(0)^{\frac{ 3 }{ 2 }}\]

the top function simplifies to \[\frac{ 3 }{ 4 } - 0\]

The bottom function simplifies to \[\frac{ 2 }{ 3 } - 0 = \frac{ 2 }{ 3 }\]

Answer 39

Alright, thats clear

Answer 40

last step!

Answer 41

I got the answer

Answer 42

step 4: subtract what we evaluated for the top function from the bottom function.

So: top= 3/4
bottom= 2/3
\[\frac{ 3 }{ 4 } - \frac{ 2 }{ 3 } = \frac{ 9 }{ 12 } - \frac{ 8 }{ 12 } = \frac{ 1 }{ 12 }\]

1/12

Answer 43

(3/4)-(2/3)

Answer 44

right

Answer 45

or .0833....

Answer 46

use fraction for integration, it's easier to work with, and in university you can't use calculators for calculus

Answer 47

So, its essentially just taking the antiderivative, then inputing top and bottom values, simplifying said values, and subtracting top value from bottom

Answer 48

You got her right there!

Answer 49

Thank you kindly! As an aside, my professor (I am in college, taking calculus 1 and 2 over the summer to avoid scheduling conflicts) is allowing calculators. Is there another method involving them?

Answer 50

Nope. You just gotta follow those steps. I'm on here almost every day so feel free to ask me for help.

Answer 51

remember to graph.

Answer 52

Thank you very much, i'll be sure to take you up on the offer. And absolutely- i'm going to attempt y=cos(t) and y=sin(t) on interval 0≤x≤pi

Answer 53

But this question is very much solved- thanks again!

Answer 54

np, if you get stuck I'll be on for a little bit. BTW a tip for you next question there is no method for finding the integral of trig functions, something you just gotta memorize.

Answer 55

Sounds good- i'll certainly be reviewing this conversation

Answer 56

=)