Integration Question (1 Viewer)

[Najji]

Find the area bounded by the curve y= sqrt(4-x^2), the x-axis and the y-axis in the first quadrant.
Thanks

 

[pokka]

Area bounded by the curve and axes is a quadrant of the circle with radius 2. So Area=(pi*2^2)/4=pi squared units.

 

[Drongoski]

You find the area without integrating - area = 1 quarter area of circle radius 2

 

[Makematics]

ok so that is the equation of a semicircle. so you know that the area in the first quadrant is a quarter of the area of a circle of radius 2.
therefore A= 1/4 pi.(2^2)=pi units^2

edit: beaten to it :/

 

[Najji]

Ohh I get it now. Thanks a lot guys.

 

[Najji]

Need help with this question; Find the area enclosed between the curve y= sqrt (4-x^2) and the line x-y+2 = 0.
I've taken the area of a circle formula into consideration this time.

 

[Najji]

Another one... Differentiate log10X
10 is the base

 

[Drongoski]

 

[Najji]

How did you get (1/ln10) (1/x)? I know when lnx gets differentiated it becomes 1/x but what about (1/ln10)?

 

[Kurosaki]

How did you get (1/ln10) (1/x)? I know when lnx gets differentiated it becomes 1/x but what about (1/ln10)?

1/1n10 is a constant.

so like when you differentiate 2x and you get 2, same principle.

 

[Makematics]

Need help with this question; Find the area enclosed between the curve y= sqrt (4-x^2) and the line x-y+2 = 0.
I've taken the area of a circle formula into consideration this time.

rightio so you draw your diagram, a semi circle and the straight line y=x+2.
clearly they will intersect at (0,2) and (2,0). From here we need to find the area of the minor segment in the 2nd quadrant. We can do this by either using the formula directly or by using the formula indirectly.
A=Area of quarter circle - area of isoscles triangle
= 1/4 pi(2)^2- 1/2 (2)(2)
=(pi-2) units^2