Integration Question help (1 Viewer)

[lisztphisto]

Could some please do this when they have time?
Find the area bounded by the curve x = -ysqaured - 5y - 6 and the y-axis.
thx every1

 

[watatank]

Could some please do this when they have time?
Find the area bounded by the curve x = -ysqaured - 5y - 6 and the y-axis.
thx every1

First factorise to find the limits....

x = -y^2 - 5y - 6

= - (y^2 + 5y + 6)

= - (y + 3)(y + 2)

If you graph that your limits will be -3 and -2

Area: Integral (lim: -3 to -2) of [ -y^2 - 5y - 6 ]

= [ {-y^3/3} - {5y/2} - 6y ] (a = -3, b = -2)

You should be able to do the rest.

 

[Forbidden.]

First factorise to find the limits....

x = -y^2 - 5y - 6

= - (y^2 + 5y + 6)

= - (y + 3)(y + 2)

If you graph that your limits will be -3 and -2

Area: Integral (lim: -3 to -2) of [ -y^2 - 5y - 6 ]

= [ {-y^3/3} - {5y/2} - 6y ] (a = -3, b = -2)

You should be able to do the rest.

YOU forgots the ² in = [ {-y^3/3} - {5y/2} - 6y ] (a = -3, b = -2)
It should be = [ {-y³/3} - {5y²/2} - 6y ] (a = -3, b = -2)

Ya, I got my intercepts the same as yours thanks.


I solve the rest k ?

= [ (-y³/3) - (5y²/2) - 6y ]

= [ (-(-3)³/3) - (5(-3)²/2) - 6(-3) ] - [ (-(-2)³/3) - (5(-2)²/2) - 6(-2) ]

= [ 9 - 22.5 + 18 ] - [ (8/3) - 10 + 12]

= [4.5] - [14/3]

A = 1/6 units²

I hopes I am right.