How do you find the definite integral of #(x) / sqrt(4 + 3x) dx # from #[0, 7]#?

1 Answer
Jim H
Oct 30, 2016

Please see the explanation section below.

Explanation:

#int_0^7 x / sqrt(4 + 3x) dx #

Let #u = 4+3x#.

This makes #du = 3 dx#, and #x = (u-4)/3# and the limits of integration become #u=4# to #u = 25#

Substitute

#1/3int_0^7 x (4 + 3x)^(-1/2) (3dx) = 1/3 int_4^25 (u-4)/3 * u^(-1/2) du#

# = 1/9 int_4^25 (u^(1/2)-4u^(-1/2)) du#

# = 1/9[2/3u^(3/2) - 8u^(1/2)]_4^25#

# = 1/9[2/3u^(3/2) - 24/3u^(1/2)]_4^25#

# = 2/27[sqrtu(u-12)]_4^25#

# = 2/27[5(13)-2(-8)]#

# = 2/27(65+16) = 2/27(81) = 6#

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