How do you find the definite integral of $\frac{x}{\sqrt{4 + 3 x}} d x$ $(x) / sqrt(4 + 3x) dx$ from $[0, 7]$ $[0, 7]$ ?

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How do you find the definite integral of $\frac{x}{\sqrt{4 + 3 x}} d x$ $(x) / sqrt(4 + 3x) dx$ from $[0, 7]$ $[0, 7]$ ?

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Answer 1 · 2016-10-30T21:45:50

Please see the explanation section below.

Explanation:

$\int_{0}^{7} \frac{x}{\sqrt{4 + 3 x}} d x$ $int_0^7 x / sqrt(4 + 3x) dx$

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

This makes $d u = 3 d x$ $du = 3 dx$ , and $x = \frac{u - 4}{3}$ $x = (u-4)/3$ and the limits of integration become $u = 4$ $u=4$ to $u = 25$ $u = 25$

Substitute

$\frac{1}{3} \int_{0}^{7} x {(4 + 3 x)}^{- \frac{1}{2}} (3 d x) = \frac{1}{3} \int_{4}^{25} \frac{u - 4}{3} \cdot u^{- \frac{1}{2}} d u$ $1/3int_0^7 x (4 + 3x)^(-1/2) (3dx) = 1/3 int_4^25 (u-4)/3 * u^(-1/2) du$

$= \frac{1}{9} \int_{4}^{25} (u^{\frac{1}{2}} - 4 u^{- \frac{1}{2}}) d u$ $= 1/9 int_4^25 (u^(1/2)-4u^(-1/2)) du$

$= \frac{1}{9} {[\frac{2}{3} u^{\frac{3}{2}} - 8 u^{\frac{1}{2}}]}_{4}^{\frac{3}{2}}$ $= 1/9[2/3u^(3/2) - 8u^(1/2)]_4^25$

$= \frac{1}{9} {[\frac{2}{3} u^{\frac{3}{2}} - \frac{24}{3} u^{\frac{1}{2}}]}_{4}^{\frac{3}{2}}$ $= 1/9[2/3u^(3/2) - 24/3u^(1/2)]_4^25$

$= \frac{2}{27} {[\sqrt{u} (u - 12)]}_{4}^{25}$ $= 2/27[sqrtu(u-12)]_4^25$

$= \frac{2}{27} [5 (13) - 2 (- 8)]$ $= 2/27[5(13)-2(-8)]$

$= \frac{2}{27} (65 + 16) = \frac{2}{27} (81) = 6$ $= 2/27(65+16) = 2/27(81) = 6$

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How do you find the definite integral of $\frac{x}{\sqrt{4 + 3 x}} d x$ $(x) / sqrt(4 + 3x) dx$ from $[0, 7]$ $[0, 7]$ ?

1 Answer

Explanation:

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

1 Answer

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How do you find the definite integral of $\frac{x}{\sqrt{4 + 3 x}} d x$ $(x) / sqrt(4 + 3x) dx$ from $[0, 7]$ $[0, 7]$ ?