Integrating Fractions (1 Viewer)

[jexca]

Hi,
I'm having trouble with a question in a prepared exam:


∫<SUP>π/4 (3sin2x – 1/4x) dx</SUP>
<SUP>0</SUP>
<SUP>(Sorry if its hard to understand - Between </SUP> π/4 and 0)

I'm fine with the 3sin2x, but I'm not sure how to integrate the 1/4.. Any help would be appreciated
Thankyou!

 

[201055]

It'll just be (-1/4) * (x^2 /2 ) = - x^2 / 8 wouldn't it?

 

[jake2.0]

It'll just be (-1/4) * (x^2 /2 ) = - x^2 / 8 wouldn't it?

If she means (1/4)*x then yes it would, but if she means 1/(4*x) then it'd just be (1/4)*log(x)

 

[201055]

If she means (1/4)*x then yes it would, but if she means 1/(4*x) then it'd just be (1/4)*log(x)

Ah yes good point...gotta be less careless....

 

[jexca]

yeah sorry, its (1/4)x

 

[jexca]

Ah I get it now, so you integrate the 1/4 and the x seperately?
Thanks

 

[webby234]

Yeah, constants are unaffected by integration, you just integrate the x and then multiply the result by the constant.

 

[jexca]

:S I understand the fraction part now, but i'm still getting the wrong answer..

When I sub π/4 and 0 in, this is what i got:

[-3cos2(π/4) - ((π/4)^2/8)] - [-3cos2(0) - 0^2/8]

= [-3cos π/2 - π^2-144] - [-3]

= 0 - π^2/144 + 3

= - π^2/144 + 3

:S

The answer is supposed to be
-π^2/128 + 3/2

 

[jake2.0]

When integrate [FONT=verdana,arial,helvetica,sans-serif][SIZE=-1][FONT=Helvetica,Arial,sans-serif]3*sin(2*x) you get[/FONT][/SIZE][/FONT][FONT=verdana,arial,helvetica,sans-serif][SIZE=-1][FONT=Helvetica,Arial,sans-serif] (-3*cos(2*x))/2[/FONT][/SIZE][/FONT]

EDIT: Also ((π/4)^2/8) = ((π^2/16)/8) = π^2/128

 

[watatank]

not sure what you done wrong but here's my working

integral (3sin2x – 1/4x) dx

= [ (3/2) * (-cos2x) ] - [ (1/4) * (1/2) x^2 ] ---- ( limits 0 to pi/4) ---

= [ (3/2) * (-cos2x) ] - [ (1/8) x^2 ] ---- ( limits 0 to pi/4) ---

= [ -3/2 * cos (pi/2) - (1/8)(pi/4)^2 ] - [ - 3/2 * cos0 - (1/8) * 0^2 ]

= [ 0 - (pi^2/128) ] - [ -(3/2) - 0 ]

= - (pi^2/128) + (3/2)