Question

adanne1990's question

Answer 1

Can someone solve these problems (calculus 2 I believe). If not then I'll try to do them myself, but it's just that im tryin to do my hw..

Answer 2

u want them all solved?

Answer 3

not me someone else does.. but im workin on my hw.. and if anyone feels like solving a couple of problems then go ahead.. adanne1990 emailed them to me askin for help.

Answer 4

o, ok

Answer 5

If it didnt say "TEST" right there on it, id be more willing to oblige ... :/

Answer 6

lol want me to rename the file?

Answer 7

and rewrite it as well ;)

Answer 8

sure..

Answer 9

lolz

Answer 10

here are a couple of problems..

Answer 11

this is the drawing ... not to scale of course

Answer 12

your integration for area will be from 1 to 5; and from 5 to 9
\[\int_{1}^{5}(x+1)-(2)\ dx+\int_{5}^{9}(11-x)-(2)\ dx\]
if I see it right

Answer 13

convert the equations to inverses:

y = x+1 ; x = y-1
y =11-x ; x = 11-y

and integrate with respect to y from 2 to 7

\[\int_{2}^{7}(y-1)-(11-y)\ dy\]

Answer 14

lol ... typoes it :)

Answer 15

\[\int_{}^{}(y-1)-(11-y)\ dy\] .. nah its good; gonna put tha tpersonaility in a time out :)

Answer 16

thanks man.. mind finishin it off?

Answer 17

y-1-11+y = 2y-12

{S} 2y-12 dy --> (2y^2)/2 -12y; y^2 -12y

area = (7)^2-12(7)-[(2)^2-12(2)]
= 49 - 84 -(4-24)
= -35 -(-20)
= -15

just caught a fauxpaux; the point above needs to be (5,6);

5+1 = 6; 11 - 5 = 6 .. which alters the integrals a little and means this set up is errored.

lets try that with y from 2 to 6

area = (6)^2-12(6)-[(2)^2-12(2)]
= 36 - 72 -(4-24)
= -36 -(-20)
= -16 <-- this negative here tells me that I just subtracted
the function in the wrong order; well, not wrong,
just backwards frm normality :)

ill revamp this for accuracy

Answer 18

A). graph is good except need the points are:
\begin{array}c
&(5,6)&\\
(1,2)&...&(9,2)\\
\end{array}

B). \(\int_{1}^{5}(x+1)-(2)dx+\int_{5}^{9}(11-x)-(2)dx\)

C). \(\int_{2}^{6}(11-y)-(y-1)dy\)

D). \(F(y)=\int_{2}^{6}(11-y)-(y-1)dy\)
\(=\int_{2}^{6}(11-y-y+1)dy\)
\(=\int_{2}^{6}(12-2y)dy\)
\(=12y-y^2\)
\begin{align}
Area &= 12(6)-(6)^2-(12(2)-(2)^2)\\
&=72-36-(24-4)\\
&=36-20\\
&=16
\end{align}

Answer 19

when we rotate about the y axis; we get a radius equal to x and a height equal to f(x); so our integration goes from 1 to 9; but becasue of the change in f(x) at 5 we need to split it up;

\[2\pi\left(\int_{1}^{5}x[(x+1)-2]dx+\int_{5}^{9}x[(11-x)-2]dx\right)\]

Answer 20

x(x-1) = x^2 - x

{S} x^2 - x dx --> x^3/3 -x^2/2

5^3/3 -5^2/2 - (1^3/3 -1^2/2)
125/3 -25/2 - 1/3 -1/2
124/3 -24/2
124(2)/6 -24(3)/6
248/6 -72/6

176/6 = 29' 1/3


x(9-x) = 9x - x^2

{S} 9x - x^2 dx --> 9x^2/2 -x^3/3

9(9)^2/2 -(9)^3/3 - (9(5)^2/2 -(5)^3/3)
729/2 -729/3 - 225/2 +125/3
504/2 -604/3
504(3)/6 -604(2)/6
1512/6 -1208/6

304/6 = 50' 2/3

50' 2/3
+ 29' 1/3
---------
79' 3/3 = 80

2pi(80) = 160pi for the volume of revolution

Answer 21

its best to dbl chk my work, I am prone to errors in this :)

Answer 22

but thats about me limit for this; good luck with it :)

Answer 23

hey assassin dou u have a facebook?