How do you solve $\int_{0}^{1} \sqrt{5 x + 4} d x$ $int_0^1 sqrt(5x+4) dx$ ?

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How do you solve $\int_{0}^{1} \sqrt{5 x + 4} d x$ $int_0^1 sqrt(5x+4) dx$ ?

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Answer 1 · 2017-01-17T00:21:18

$\frac{38}{15}$ $38/15$

Explanation:

We need to use a u-substitution to solve this question.

STEP 1: Identify the u
$u = 5 x + 4$ $u = 5x+4$

STEP 2: Find du
$d u = 5 d x$ $du = 5 dx$

STEP 3: Change the x-bounds to u-bounds
$x = 0 \to u = 5 (0) + 4 = 4$ $x = 0 -> u = 5(0)+4 = 4$
$x = 1 \to u = 5 (1) + 4 = 9$ $x = 1 -> u = 5(1)+4 = 9$

STEP 4: Do the u-substitution
$\int_{0}^{1} \sqrt{5} x + 4 d x$ $int_0^1 sqrt 5x+4 dx$
$= \int_{4}^{9} \sqrt{u} \cdot \frac{1}{5} d u$ $=int_4^9 sqrt(u) * 1/5 du$
remember: we found $d u = 5 d x$ $du = 5dx$ , so if we solve for $d x$ $dx$ we get $d x = \frac{1}{5} d u$ $dx=1/5 du$
$= \frac{1}{5} \int_{4}^{9} \sqrt{u} d u$ $=1/5 int_4^9 sqrt(u) du$
$= \frac{1}{5} {[\frac{2}{3} u^{\frac{3}{2}}]}_{4}^{\frac{3}{2}}$ $=1/5 [2/3 u^(3/2)]_4^9$
$= \frac{2}{15} {[u^{\frac{3}{2}}]}_{4}^{\frac{3}{2}}$ $=2/15[u^(3/2)]_4^9$
$= \frac{2}{15} (3^{3} - 2^{3})$ $=2/15(3^3 - 2^3)$
$= \frac{2}{15} (19)$ $=2/15(19)$
$= \frac{38}{15}$ $=38/15$

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How do you solve $\int_{0}^{1} \sqrt{5 x + 4} d x$ $int_0^1 sqrt(5x+4) dx$ ?

1 Answer

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How do you solve $\int_{0}^{1} \sqrt{5 x + 4} d x$ $int_0^1 sqrt(5x+4) dx$ ?