Question

can somone help me find the intersection of these two function, y=x^4 and y=8x

Answer 1

i did x^4=8x and got

Answer 2

x(x-2)(x^2+2x+4)

Answer 3

x^4=8x

(0,0) is obivous

Answer 4

x=2

Answer 5

right of course

Answer 6

so i got limits of integration at [0,2], does that qudratic contribute anything

Answer 7

No

Answer 8

x^4=8x , x^4 - 8x=0 , x(x^3-8)=0 , x=0 or x^3 - 8 =0 , x=0 or x^3=8 , x=0 or x=2

Answer 9

get that mathcruncher?

Answer 10

huh i yeah i got it

Answer 11

would the 2 be the upper limit of integration and the 0 the lower limit

Answer 12

Yes, you just need interesection point

Answer 13

This is a quartic and therefore has 4 roots.

by inspection, 0 and 2 are real, the other 2 are complex.

Answer 14

right, but when i evluated the intergal for the are i got -48/5, the book says i should have gotten +48/5

Answer 15

x^4 is positive for all x.

Answer 16

so should i just recongnize that i take the answer as positive

Answer 17

or is there somthing i am missing

Answer 18

cause when i graph these function, i see that the area bounded by these functions is in the first qudrant

Answer 19

I don't know what integral u are referring to.

Since x^4 is positive, any area under it must be positive.

Answer 20

you know the integral formed by those two functions

Answer 21

evaluated [0,2]

Answer 22

You are not making any sense, the line 8x is above the curve between 0 and 2.

Answer 23

right, so i am looking to evluate the area bounded by those two curves

Answer 24

are u saying u want the area between the curves?

Answer 25

the area of the region enclosed by the line and curve

Answer 26

Just calculate the area of the triangle and subtract the integral of 4x^2 between 0 and 2.

Answer 27

1/2 *h*b

Answer 28

It's 16 obviously.

Answer 29

right

Answer 30

its obviously not 16

Answer 31

what are you talking about estudier

Answer 32

i evluated the integral and got -48/5, but the answer is positive 48/5

Answer 33

\[y=x^4, y=8x\]
To find where these two function intersect, set them equal to each other:
\[x^4=8x \rightarrow \frac{x^4}{x}=8 \rightarrow x^3=8 \rightarrow \sqrt[3]{x^3}=\sqrt[3]{8} \rightarrow x=2\]

Answer 34

very good, now take the integral and evluate the are

Answer 35

can you tell me what you get

Answer 36

cause my answer is alittle off i guess

Answer 37

are you trying to find the point where they intersect on the intersection of the areas under both functions?

Answer 38

The 16 is the area of the triangle, I assumed that's what you meant by 1/2bh.

Now u need to subtract the area under x^4 between 0 and 2.

Answer 39

@nadeem we have already found the intersection points, he want s the area bounded by the line and the curve.

Answer 40

so, its gonna be 16-35/5

Answer 41

Which is the triangle formed by the line and the x-axis of area 16 less the area under the curve x^4 between 0 and 2.

Answer 42

I haven't done the integration, by eye it looks to be about 8 or so.

Answer 43

okay, but isnt the the integral from 0 to 2, of x^4-8x

Answer 44

Why do you want to do that?

Answer 45

find the area

Answer 46

U already found it, I can see u wrote it up above, 16-7 = 9.

Answer 47

That's the simplest easiest fastest way to get the area.

Answer 48

but thats not the answer

Answer 49

k:
\[\int\limits_{0}^{2} 8x-x^4dx = \int\limits_{0}^{2} 8xdx- \int\limits_{0}^{2} x^4dx= \left[ 4x^2 \right]_{0}^{2} - \left[ \frac{x^5}{5} \right]_{0}^{2}= 16-\frac{32}{5}= \frac{48}{5}\]

Answer 50

thats what im looking for

Answer 51

how come the function 8x goes before the x^4

Answer 52

oh itst the upper curve

Answer 53

because the upper function is 8x and the lover function is x^4, on these kinds of integrals graphing the functions helps tremendously, and it puts everything in perspective as well.

Answer 54

thanks

Answer 55

sure thing

Answer 56

16 - int x4 [0,2]^= 16 - x^5/5 -> 16- 32/5 = 48/5 QED.