Help on intergration (1 Viewer)

[kittyful]

Find the area enclosed by the curve y= root 3x-5, the axis and the lines y = 2 and y =3

 

[Bored Of Fail 3]

Find the area enclosed by the curve y= root 3x-5, the axis and the lines y = 2 and y =3

thats easy

 

[stevey6404]

Make x the subject of the equation by squaring the root & y and then rearrange etc.

and then integrate from there with respect to dy.

 

[Bored Of Fail 3]

y= sqrt (3x-5)
y^2 = 3x -5
3x = ( 5+y^2)
x= (1/3) ( 5+y^2)

therefore area = integral x dy

= (1/3) integral ( 5+y^2) dy [ from 2..3]
= (1/3) [ 5y + y^3 /3 ] ( 2..3)
= (1/3) [ ( 15 + 9) - ( 10 +8/3 ) ] = 34/9 units^2

 

[kittyful]

Thanks guys for your help i just really didn't get how the 1/3 was factored out of the thing, ... do you always have to factor out the 1/3 is it possible to do the working without factoring out the 1/3 ?

 

[Bored Of Fail 3]

Thanks guys for your help i just really didn't get how the 1/3 was factored out of the thing, ... do you always have to factor out the 1/3 is it possible to do the working without factoring out the 1/3 ?

its a constant , it has nothing to do with the integral , it does not depend on y ( the variable we are integrating with respect to).

It can be factored out . you could leave it inside but its muccccch easier to take constants out the front of an integral sign

 

[kittyful]

oh right thanks i understand this much better now

I was wondering about this question x= - y^2 -5y - 6 , i drew this out its a sideways parabola, is the intergral thing from 6 to -1 ?

 

[Bored Of Fail 3]

whats the exact question?

 

[kittyful]

Find the area bounded by the curve x= - y^2 -5y - 6 and the y axis

 

[Bored Of Fail 3]

Find the area bounded by the curve x= - y^2 -5y - 6 and the y axis

according to wolfram (i was to lazy to complete the square and graph by hand )


it is not a sideways parabola

http://www.wolframalpha.com/input/?i=x=+-+y^2+-5y+-+6

there is no well defined area between this curve and the y axis

do they give limits, they should

 

[Bored Of Fail 3]

or they could possibly mean

"Find the area bounded by the curve x= - y^2 -5y - 6 and the y axis AND the x axis "

 

[kittyful]

nope they didn't give limits .. just the y axis .

 

[Bored Of Fail 3]

do they give an answer?

I think its | integral ( -y^2 -5y -6 ) dy between [y=0 and y=6] |

 

[kittyful]

oh and ummm just one help on one more question "find area bounded by the curve y=1 and the curve y=x^2" with this i summed the equations together but i am not quite getting the answer.

 

[kittyful]

do they give an answer?

I think its | integral ( -y^2 -5y -6 ) dy between [y=0 and y=6] |

the answer is 1/6 units sq

 

[Bored Of Fail 3]

well this is an example of area between curves.

so first we find intersections by solving simultaneously

they both equal y, so set them equal to each other

x^2 =1
x^2 -1 =0
(x-1)(x+1) =0

x=-1, x=1

now, if you draw a diagram you will see that y=1 is above y=x^2 in the interval -1<= x<=1


so area = integral (top y - bottom y) dx [ between limits of x=-1 and x=1 ]

so integral (1-x^2) dx [ -1..1]

= ( x -x^3 / 3 ) [ -1..1]
= ( 1 -1/3 ) - ( -1 - (-1/3) ) = 2/3 - ( -2/3) = 4/3 units^2

 

[kittyful]

thank you so much, i did exactly the same working a you got up there, maybe its how i used the calculator thanks i appreciate it

 

[deterministic]

Find the area bounded by the curve x= - y^2 -5y - 6 and the y axis

diagram looks like the attachment. Area you want is in red.

So answer is

 

[Bored Of Fail 3]

View attachment 22682
diagram looks like the attachment. Area you want is in red.

So answer is

lol wolfram let me down, doesnt label the axes on the graphs :|

 

[kittyful]

oh so its a sideways parabola ?