Integration Q (1 Viewer)

[shaon0]

Find the area bounded by y=((x^2)/4) -2 and y=x+1.
I have got the answer but i don't know whether it is correct.
100/3 is my answer. Could someone please verify this?

 

[shaon0]

How do you integrate:
(1-x)/(x^2)

 

[pLuvia]

Just split the fraction into 2 i.e. 1/x²-x/x² then integrate

 

[shaon0]

Just split the fraction into 2 i.e. 1/x²-x/x² then integrate

Thanks, i didn't think of that.

 

[Aerath]

Find the area bounded by y=((x^2)/4) -2 and y=x+1.
I have got the answer but i don't know whether it is correct.
100/3 is my answer. Could someone please verify this?

I got 27/4. But I'm not sure.

 

[shaon0]

I have another question:
The area below y=1/x in the first quadrant between x=1 and x=4 is rotated about the x axis. What is the volume of the generated solid?
I have done this question and i get 15(pi)/16 but the answer is 3(pi)/4. I can't find the fault in my working.
Thanks

 

[tommykins]

y² = 1/x²

v = pi int x^-2 dx 4->1

= pi [ -x^-1 ] 4->1
= pi [ -1/4 - [-1/1] ]
= pi [ -1/4 + 1 ]
= 3pi/4

 

[shaon0]

y² = 1/x²

v = pi int x^-2 dx 4->1

= pi [ -x^-1 ] 4->1
= pi [ -1/4 - [-1/1] ]
= pi [ -1/4 + 1 ]
= 3pi/4

....I forgot to integrate my 1/x^2. lol
Thanks for your help

 

[shaon0]

Integration Q:
The region in the first quadrant bounded by the graphs of f(x)=(x^3)/8 and g(x)=2x is rotated around the y axis. Find the volume of the solid found.
I think i have followed the right method but i have got the wrong answer.
Answer: 107.223 c.cm

 

[lolokay]

x³ - 16x = 0
x = 0, 4

pi Int (x⁶/64) - 4x²
= pi [x⁷/(64*7) - 4/3 x³]4,0
=~ 153.19

is that what you got?

 

[shaon0]

x³ - 16x = 0
x = 0, 4

pi Int (x⁶/64) - 4x²
= pi [x⁷/(64*7) - 4/3 x³]4,0
=~ 153.19

is that what you got?

No...around the y-axis. so you sub the x-values into one of the equations.
And you get the y limit which is 0,8....i got.
(32pi/3)-(384pi/5)

 

[lolokay]

oh didn't see that bit

by rearranging the equations you get

x² = y²/4 and x² = 4y^2/3
with intercepts y=0,8

pi Int y²/4 - 4y^2/3
= pi(y³/12 - 12/5 y^5/3) for y=8
= 128pi/3 - 384pi/5
= -107.233

so volume = 107.233

your answer was close. got 32 instead of 128

 

[shaon0]

oh didn't see that bit

by rearranging the equations you get

x² = y²/4 and x² = 4y^2/3
with intercepts y=0,8

pi Int y²/4 - 4y^2/3
= pi(y³/12 - 12/5 y^5/3) for y=8
= 128pi/3 - 384pi/5
= -107.233

so volume = 107.233

your answer was close. got 32 instead of 128

Yea i figured it out before i saw your solution. I forgot to put a 2^2 in front of one of 32 when integrating for some reason.
Thanks for your help
btw, is this a year 12 topic or year 11?

 

[lolokay]

according to the syllabus it's a year 12 topic