HSC PAPER 1993 3 c ii (2 Viewers)

[bemer]

how do you work out such a question
i couldnt understand the solution in successs one
any1 ??

 

[addikaye03]

how do you work out such a question
i couldnt understand the solution in successs one
any1 ??

well you need to have done trig functions for a start.

HSC 1993 Q3cii) INT (0-->pi/2) sin2x dx

[-1/2cos2x](pi/2-->0)

[-1/2cospi]-[-1/2]

=1

 

[bemer]

i get -.5 cos2x

but i don noe how to sub in pi/2 and 0

 

[addikaye03]

i get -.5 cos2x

but i don noe how to sub in pi/2 and 0

sorry bout the delay. Well when an integral is "definite" is has LIMITS. These are the values in which the function's area is determined between. So to find the area between pi/2 and 0, you sub in the larger limit then subtract the smaller limit from that value.

A=int(b-->a) x dx ( for example)

=[x^2]( from b-->a) ( this function is definite and therefore there is no constant at the end)

=[b^2]-[a^2]

if ya get me... here b>a by the way, so its the larger limit-smaller limit

 

[bemer]

sorry bout the delay. Well when an integral is "definite" is has LIMITS. These are the values in which the function's area is determined between. So to find the area between pi/2 and 0, you sub in the larger limit then subtract the smaller limit from that value.

A=int(b-->a) x dx ( for example)

=[x^2]( from b-->a) ( this function is definite and therefore there is no constant at the end)

=[b^2]-[a^2]

if ya get me... here b>a by the way, so its the larger limit-smaller limit

yeh i got u
soo i do {-.5cos2(pie on 2)} minus {-.5cos2(0)
wich equals -.49992485749 minus -.5
wich equals zero
lol wat did i do rong

 

[addikaye03]

yeh i got u
soo i do {-.5cos2(pie on 2)} minus {-.5cos2(0)
wich equals -.49992485749 minus -.5
wich equals zero
lol wat did i do rong

i suspect u dont have ur calculator is radians mode which is required when dealing with trig functions.

[-1/2cos2x](pi/2-->0)

[-1/2cos2(pi/2)]-[-1/2cos2(0)]

By graphical inspection =1 OR:

[-1/2cospi]-[-1/2] (just simplified it)

-1/2( cospi-1)

so -1/2(-1-1)

-1/2(-2)=1

Do u kinda get that? i think ur prob was the radians thing.