Question

Determine the area of the enclosed by
y=3x2
and
y=6x


and

Determine the area enclosed by the graphs of
y=x3
and
y=4x
for
x≥0

what is really the difference between the two questions and how are they solved

Answer 1

this is not double integral no?

Answer 2

what is the interval for the first one

Answer 3

nno intervals given but with the interception, we can find the intervals

Answer 4

yes correct! what did you find from the graph

Answer 5

wish is x1=0 to x2=2 for the first one

Answer 6

so \(\large \int_{0}^{2}(3x^2-6x)dx\)

Answer 7

area enclosed by the curves

Answer 8

yes

Answer 9

Is 6x>3x^2 on [0,2]?

Answer 10

actually that depend on which is upper and lower!

Answer 11

so could be 6x-3x^2

Answer 12

It is 6x-3x^2 since 6x>3x^2 on [0,2]

Answer 13

my result gave 4.... but when i tried solving the second question, i got different result

Answer 14

yes @ myininaya

Answer 15

yeah right it is

Answer 16

can you help with the second

Answer 17

well it is the same process! what are your boundaries

Answer 18

no interval

Answer 19

well there must be

Answer 20

[0,?]

Answer 21

0,2

Answer 22

am just lost in this.....
the answer to the first one is 8 and to the second, it is 4.....
please enplane,,,,, what is the difference

Answer 23

so 4x>=x^3 in the interval [0,2]

Answer 24

this is what you did

Answer 25

yes

Answer 26

and you got 4

Answer 27

yes after integrating and substituting x1 and x2

Answer 28

do the regions looks similar?

Answer 29

that is what i do not get

Answer 30

i haven't done the graphs so don't know

Answer 31

ok. thanks for your help any ways

Answer 32

you are welcome!
i forgot this stuff lol

Answer 33

@GIL.ojei Are you saying you are having trouble with the integration?

Answer 34

well, yes but mostly,how to know the region

Answer 35

Well it is obvious to me you can find the intersections pretty easily...You have been able to answer that question each time.

Answer 36

If you want to know what function is larger than the other, you could test the interval in between each intersections by pluggin in a number

Answer 37

ok

Answer 38

i think he didn't get which function is larger than the other in the interval

Answer 39

Say I want to find the area bounded by y=x^2 and y=x.
So x^2=x when x=1 or x=0 .
We know x>x^2 because 1/2>1/4 (when you plug in 1/2)

so we know we are going to have to do
int(x-x^2,0..1)

Answer 40

you can see it clearly from the graph

Answer 41

You could also do this with a visual as @xaspprachesinfinity suggested
However, if you want to skip the visual you can. But a visual does help.

Answer 42

@xapproachesinfinity ,i do not know how to draw the graph

Answer 43

@GIL.ojei Do you understand what I was saying?

You know how to find the intersections for y=x^3 and y=4x for x>=0

you seen that x^3=4x when x=-2,0,2
x=-2 does matter since x>=0


so we are only looking as a visual between 0 and 2 for the graphs y=x^3 and y=4x
To see which is bigger on that interval, simply plug in a number between 0 and 2 like 1.
Let f(x)=x^3 and g(x)=4x
f(1)=1
g(1)=4
g>f since 4>1
so 4x>x^3 on the interval [0,2]

Answer 44

you draw both graphs you will see that they intersect somewhere
that the same thing as saying
3x^2=6x to find the intersections

Answer 45

so you do big-small

Answer 46

doesn't matter* (i didn't mean to say does matter)

Answer 47

i find graphs to be much more easier to understand!
Ate any rate, the result is the same^_^

Answer 48

Graphs are very useful.

Answer 49

so if x1 > x2 what becomes the region @myininaya

Answer 50

So I would learn how to graph these basic algebraic graphs.

Answer 51

what is x1 and x2?

Answer 52

the intervals

Answer 53

If the function f is bigger than the function g on the interval [x_1,x_2], then you setup the integral as follows:
\[\int\limits_{x_1}^{x_2}(f(x)-g(x)) dx \]