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

1 Answer
Jim H
Oct 30, 2016

Please see the explanation section below.

Explanation:

int_0^7 x / sqrt(4 + 3x) dx ∫70x√4+3xdx

Let u = 4+3xu=4+3x.

This makes du = 3 dxdu=3dx, and x = (u-4)/3x=u−43 and the limits of integration become u=4u=4 to u = 25u=25

Substitute

1/3int_0^7 x (4 + 3x)^(-1/2) (3dx) = 1/3 int_4^25 (u-4)/3 * u^(-1/2) du13∫70x(4+3x)−12(3dx)=13∫254u−43⋅u−12du

= 1/9 int_4^25 (u^(1/2)-4u^(-1/2)) du=19∫254(u12−4u−12)du

= 1/9[2/3u^(3/2) - 8u^(1/2)]_4^25=19[23u32−8u12]254

= 1/9[2/3u^(3/2) - 24/3u^(1/2)]_4^25=19[23u32−243u12]254

= 2/27[sqrtu(u-12)]_4^25=227[√u(u−12)]254

= 2/27[5(13)-2(-8)]=227[5(13)−2(−8)]

= 2/27(65+16) = 2/27(81) = 6=227(65+16)=227(81)=6

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