Area Problem (1 Viewer)

[dawso]

Find the exact area enclosed between the curve y=e^(3x) - 2, the x axis and the lines x=0 and x=1

i swear the texbook is wrong, wat do u guys get>

 

[Pace_T]

1/3(e^3 -7) u^2
is that what u got?

 

[FinalFantasy]

Find the exact area enclosed between the curve y=e^(3x) - 2, the x axis and the lines x=0 and x=1

i swear the texbook is wrong, wat do u guys get>

area=int. e^(3x)-2 dx from 0 to 1
=(1\3)e^(3x) -2x from 0 to 1
applying da things, area=(1\3)e^3-2-(1\3)=e³\3-7\3
=(e³-7)\3 units squared

 

[Pace_T]

hey how did u get those little things at the top (the powers)??
cuase my way is confusing lol

 

[FinalFantasy]

alt 0178 for squared
alt 0179 for cube

 

[dawso]

nah guys ur way off for starters, the graph crosses the x axis in the interval so u have to get the interval between 0-(ln2)/3 and between (ln2)/3 and 1, but they just get a weird answer....

 

[dawso]

no........the other area is still enclosed by the curve and x axis....its all the area up 2 the line x=1 hm......y did i post this question in the 2unit forum?? can a mod please put it in 3 or 4 for me cause they 2unit people are struggling wit it???

 

[Pace_T]

is this a 2 unit question???

 

[FinalFantasy]

oops lol, juz started doing the normal area thing without looking

 

[FinalFantasy]

Find the exact area enclosed between the curve y=e^(3x) - 2, the x axis and the lines x=0 and x=1

at y=0, 2=e^(3x)
ln2=3x
x=(ln2)\3

area=-int. (e^(3x)-2)dx from 0 to (ln2)\3+int. (e^(3x)-2) dx from (ln2)\3 to 1
=-[(1\3)e^(3x)-2x] from 0 to (ln2)\3+[(1\3)e^(3x)-2x] from (ln2)\3 to 1
=-[(1\3)e^(ln2)-2(ln2)\3-(1\3)]+[(1\3)e^(ln2)-2(ln2)\3-(e³\3-2)]
=-2\3+2(ln2)\3+(1\3)+2\3-2(ln2)\3-e³\3+2
=1\3-e³\3+2
=(7-e³)\3 units squared

is dat rite?

 

[Trev]

Find the exact area enclosed between the curve y=e^(3x) - 2, the x axis and the lines x=0 and x=1

at y=0, 2=e^(3x)
ln2=3x
x=(ln2)\3

area=-int. (e^(3x)-2)dx from 0 to (ln2)\3+int. (e^(3x)-2) dx from (ln2)\3 to 1
=-[(1\3)e^(3x)-2x] from 0 to (ln2)\3+[(1\3)e^(3x)-2x] from (ln2)\3 to 1
=-[(1\3)e^(ln2)-2(ln2)\3-(1\3)]+[(1\3)e^(ln2)-2(ln2)\3-(e³\3-2)]
=-2\3+2(ln2)\3+(1\3)+2\3-2(ln2)\3-e³\3+2
=1\3-e³\3+2
=(7-e³)\3 units squared

is dat rite?

No, sorry lol.
You have to take into consideration that it crosses the x-axis around x=.22 (just used Graphmatica to see this), so you will have to find the area between x=0 and x=~.22 (below the x-axis) and add this to the area between x=~.22 and x=1 (above the x-axis)
If you tested the answer you got FinalFantasy [(7-e³)\3] you would see that it is negative, and area can't be of that nature!

 

[FinalFantasy]

No, sorry lol.
You have to take into consideration that it crosses the x-axis around x=.22 (just used Graphmatica to see this), so you will have to find the area between x=0 and x=~.22 (below the x-axis) and add this to the area between x=~.22 and x=1 (above the x-axis)
If you tested the answer you got FinalFantasy [(7-e³)\3] you would see that it is negative, and area can't be of that nature!

"so you will have to find the area between x=0 and x=~.22 (below the x-axis) and add this to the area between x=~.22 and x=1 (above the x-axis)"
i did dat
.22=my (ln2)\3
i got area below(dats y i put a negative in front of the integral)
den i added the positive area...
but i dunno y i still get negative after doing dat lol

 

[FinalFantasy]

oh i applied terminals wrong way around on da last bit LoL!

here it should be:
area=-int. (e^(3x)-2)dx from 0 to (ln2)\3+int. (e^(3x)-2) dx from (ln2)\3 to 1
=-[(1\3)e^(3x)-2x] from 0 to (ln2)\3+[(1\3)e^(3x)-2x] from (ln2)\3 to 1
=-[(1\3)e^(ln2)-2(ln2)\3-(1\3)]+[(1\3)e³-2-(1\3)e^(ln2)+2(ln2)\3]
=-2\3+2(ln2)\3+(1\3)+e³\3-2-2\3+2(ln2)\3
=-3+4(ln2)\3+e³\3 units squared

 

[Pace_T]

ok umm i split it up into 2 different areas which was divided by x= ~.22
and i got A= 1/3 * (e^3-7) u²
which is positive.

 

[FinalFantasy]

Using the intervals dawso mentioned; here is my answer:
(ln2)/3 (ln2)/3
∫ (e^3x-2)dx + ∫ (e^3x-2)dx
0 0

[(e^3x)/3 - 2x] (ln2)/3 | 0 + [(e^3x)/3 - 2x] 1 | (ln2)/3
[(e^ln2)/3 - (2ln2)/3 - 1/3] + [(e³)/3 - 2 - (e^ln2)/3 + (2ln2)/3]
=e³/3 - 7/3
Which I am quite sure is the correct answer... Integration is great!

^^^EDIT: Sorry about the screwed up nature of the working in the beginning, hopefully you get the drift!

dats the thing i got up there at the very first post

 

[FinalFantasy]

but e³/3 - 7/3 is different from my recent one lol
i believe my recent one is right!

area=-int. (e^(3x)-2)dx from 0 to (ln2)\3+int. (e^(3x)-2) dx from (ln2)\3 to 1
=-[(1\3)e^(3x)-2x] from 0 to (ln2)\3+[(1\3)e^(3x)-2x] from (ln2)\3 to 1
=-[(1\3)e^(ln2)-2(ln2)\3-(1\3)]+[(1\3)e³-2-(1\3)e^(ln2)+2(ln2)\3]
=-2\3+2(ln2)\3+(1\3)+e³\3-2-2\3+2(ln2)\3
=-3+4(ln2)\3+e³\3 units squared

 

[dawso]

yeah thats wat i got 2, hm.........

 

[FinalFantasy]

yeah thats wat i got 2, hm.........

wat does da answer say den?