Asked by Jay
evaluate the definite integrals
1
∫ root(1+3x) dx
0
1
∫ root(1+3x) dx
0
Answers
Answered by
Reiny
= [(2/3)(1/3)(1 + 3x)^(3/2)] from 0 to 1
=[(2/9)(1 + 3x)^(3/2)] from 0 to 1
= (2/9)(4^(3/2) - (2/9)(1^(3/2))
= (2/9)(8) - (2/9)
= 14/9
=[(2/9)(1 + 3x)^(3/2)] from 0 to 1
= (2/9)(4^(3/2) - (2/9)(1^(3/2))
= (2/9)(8) - (2/9)
= 14/9
Answered by
Damon
1
∫ (1+3x)^.5 dx
0
let z = (1+3x)
dz = 3 dx so dx = (1/3)dz
if x = 0, z = 1
if x = 1, z = 4
so
4
∫ (z)^.5 dz/3
1
is
(1/4.5) z^1.5 at z = 4 - at z = 1
∫ (1+3x)^.5 dx
0
let z = (1+3x)
dz = 3 dx so dx = (1/3)dz
if x = 0, z = 1
if x = 1, z = 4
so
4
∫ (z)^.5 dz/3
1
is
(1/4.5) z^1.5 at z = 4 - at z = 1
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