2007-09-01 01:53:06 · 5 answers · asked by Michael H 2 in Science & Mathematics ➔ Mathematics
times 2 Pi on the interval 0, 1
2007-09-01 02:04:45 · update #1
I = ∫ y^(4/3) - y ³ dy
I = (3 / 7) y^(7/3) - y^4 / 4 + C
I see limits 0 to 1 have been added:-
I = (3 / 7) - 1 / 4
I = 12 / 28 - 7 / 28
I = 5 / 28
I now notice that a 2π has suddenly appeared!
I = 10π / 28 = 5π / 14
2007-09-05 01:03:14 · answer #1 · answered by Como 7 · 0⤊ 0⤋
Int (y^4/3 - y^3) dy
= Int (y^4/3 dy) - Int (y^3 dy)
= y^5/15 - y^4/4 + c
= (1/15)y^5 - (1/4)y^4 + C
2007-09-01 02:02:59 · answer #2 · answered by vlee1225 6 · 0⤊ 0⤋
integ (y^4/3 - y^3 ) dy
= y^(4/3 + 1) / (4/3 + 1) - y^4 /4 + c
= y^ 7/3 / (7/3) - y^4 /4 + c
= 3/7 y^ 7/3 - y^4 /4 + c
2007-09-01 02:12:17 · answer #3 · answered by CPUcate 6 · 0⤊ 0⤋
â (y^(4/3) - y^3 dy =
â¡
â (y^(4/3) dy - â y^3 dy =
â¡.....................â¡
power rule
(1/(1+(4/3)))*y^(1+(4/3))-(1/(3+1)*y^(3+1) =
simplify
(3/7)*y^(7/3) - (1/4)*y^4
using the numberpad press
alt 244
and
alt 245
for integrands
2007-09-01 02:08:15 · answer #4 · answered by Mitchell 5 · 0⤊ 0⤋
â«(x^4/3 - x^3)dx
1/3â«x^4 dx -â«x^3dx
1/3 x^5/5 - x^4/4
1/15 x^5 - 1/4 x^4
x^4 (x/15 - 1/4)
x^4 (4x/60 - 15/60)
x^4 (4x-15) / 60
2007-09-01 02:09:33 · answer #5 · answered by GPC 3 · 0⤊ 0⤋