how do i integrate square root (4-x^2), with lower limit 0, upper limit pi/2, w/o using substitution? and if i wna use substitution, how do i noe which to choose? y is my tcher's substitution 2sin diter?
2007-12-20 23:50:26 · 3 answers · asked by blahhs 1 in Science & Mathematics ➔ Mathematics
if you want to integrate that without the substitution...
then realize that the integral with limits is just the area of the region bounded by the curve and coordinate axis...
now ... y = √(4 - x²) is a semicircle of radius 2
... but why are the limits 0 & π/2... i think these are for the radian measure...
if it would be the radian measure... then the area required would be the area of the quarter circle of radius 2.
thus (1/4) 4π = π
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2007-12-21 01:11:09 · answer #1 · answered by Alam Ko Iyan 7 · 0⤊ 0⤋
I = â« (2² - x²)^(1/2) dx between 0 and Ï/2
This is a STANDARD INTEGRAL in text books:-
I = sin ^(-1) (x/2) between 0 and Ï/2
I = sin^(-1) (Ï/4)
2007-12-21 09:24:23 · answer #2 · answered by Como 7 · 1⤊ 0⤋
4x-(x^3)/3=6.28-(3.86/3)-0=6.28-1.29=4.99
2007-12-21 08:01:33 · answer #3 · answered by Anonymous · 0⤊ 1⤋