Integration Q's (1 Viewer)

[mtsmahia]

1)
Find the area bounded by y= square root of ( 4- x^2) , the x-axis and the y-axis in the 1st Quadrant.

2)
Find the area bounded by the x-axis, the curve y=x^3 and the lines x=-a and x=+a


THANKS !

 

[shaon0]

1)
A= pi.(2)^2/4 [Area of a quarter of a circle as sqrt(4-x^2) is a semicircle but area is only defined as 1st quadrant]
=pi

2) S {x=-a to x=a} x^3 dx = (1/4) [x^4] {x=-a to x=a}
A = (1/4) (a^4+(-a)^4)
= (1/4) (a^4+a^4)
= a^4/2

 

[addikaye03]

2)
Find the area bounded by the x-axis, the curve y=x^3 and the lines x=-a and x=+a

f(-x)=-(x^3)=-f(x). Therefore rotational symmetry about origin, hence A=0.

 

[biopia]

These are two important rules to learn as apart of the integration topic.

Get used to recognising that a formula in that form is a semicircle.
y=√4-x²
y²=4-x²
x²+y²=4
If you get used to 'seeing' that relationship, then you won't have to traditionally integrate (which I am sure yo know is a pain because of that square root sign hehe)

The second is a rule that you can be asked to prove. It's not that hard as long as you know you will end up with an integral of 0 at the end. It only ever works like this if the function is odd and the limits are the same magnitude but have the opposite sign.

 

[hello-there]

Q1) A of circle = pi.r^2
the graph is a semi circle x-int at x=2,-2
thus, A in 1st quad=a quater of a full circle
A=pi.2^2 /0.24

Q2)An alt. method.

Since the limits are at a & -a therefore the integral of an odd function with limits a and -a is 0
also the integral of an even function " ........................" is the same as 2x the integral of the even function with limits 0 to a OR -a to 0.

 

[hscishard]

Wow, Wow, Wow, he did ask for the AREA didn't he? So that means using the the integral of odd functions with bounds +a and -a = 0 doesn't work. We use the 2 times Integral x^3, bounds being +a and 0.

: D sorry if im an ignorant bastard that doesn't know his maths if im wrong. We learn from mistakes, don't we?

 

[shaon0]

Wow, Wow, Wow, he did ask for the AREA didn't he? So that means using the the integral of odd functions with bounds +a and -a = 0 doesn't work. We use the 2 times Integral x^3, bounds being +a and 0.

: D sorry if im an ignorant bastard that doesn't know his maths if im wrong. We learn from mistakes, don't we?

oh damn, yeah sorry.

 

[addikaye03]

oh damn, yeah sorry.

I apologise also, how clumsy. Well integrate from a->0 + 0->-a

 

[shaon0]

I apologise also, how clumsy. Well integrate from a->0 + 0->-a

Looks like we both made the same mistake lol

 

[biopia]

Looks like we both made the same mistake lol

Same lol...

 

[hscishard]

Beaten by a Year 11 student . Now im just showing off.
So..Integral of 2 times x^3 bounds a and 0
= x^4 / 4.
= 2 times A^4 / 4 - 0
= a^4/2

I think thats right