Evaluate definite integral
(y+4/sqrty) dy
upper = 4
lower = 1
2007-08-10 00:14:23 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
If you have written this correctly, as
int(y + (4/sqrt(y)), dy), it evaluates to
(1/2)*y^2 + 8*sqrt(y),
then plug your limits in
=(1/2)*(16-1) + 8*(2-1)
=(15/2) + 8
=31/2
2007-08-10 01:45:53 · answer #1 · answered by Not Eddie Money 3 · 0⤊ 0⤋
u-sub
u = sqrt(y)
u = y^(1/2)
du = (1/2)*y^(-1/2)dy
2du = 1/sqrt(y)dy
change limits, upper = sqrt(4), lower = 1
integral (2u^2 + 8) du
sum rule: integral 2u^2 du + integral 8 du
2 * integral u^2 du + integral 8 du
2 * (1/3) u^3 + 8*u
answer: 64/3 - 26/3 = 38/3
2007-08-10 00:40:35 · answer #2 · answered by wilmer 5 · 0⤊ 0⤋
"Not Eddie Money" is correct.
2007-08-11 01:51:34 · answer #3 · answered by Tony 7 · 0⤊ 0⤋