What is the antiderivative of $6 \sqrt{x}$ $6sqrtx$ ?

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What is the antiderivative of $6 \sqrt{x}$ $6sqrtx$ ?

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Answer 1 · 2016-04-16T21:47:43

$\int 6 \sqrt{x} d x = 4 x^{\frac{3}{2}} + C$ $int6sqrt(x)dx = 4x^(3/2)+C$

Explanation:

In general, we have $\int x^{k} d x = \frac{x^{k + 1}}{k + 1} + C$ $intx^kdx = x^(k+1)/(k+1)+C$ for $k \neq - 1$ $k!=-1$ . Putting that together with $\int k f (x) d x = k \int f (x) d x$ $intkf(x)dx = kintf(x)dx$ we have:

$\int 6 \sqrt{x} d x = 6 \int x^{\frac{1}{2}} d x = 6 \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + C = 4 x^{\frac{3}{2}} + C$ $int6sqrt(x)dx = 6intx^(1/2)dx = 6x^(3/2)/(3/2)+C = 4x^(3/2)+C$

Answer 2 · 2016-04-16T21:49:21

$4 x^{\frac{3}{2}} + C$ $4x^(3/2)+C$

Explanation:

We want to find

$\int 6 \sqrt{x} d x$ $int6sqrtxdx$

Which, since $6$ $6$ is a multiplicative constant, equals

$= 6 \int \sqrt{x} d x$ $=6intsqrtxdx$

Write $\sqrt{x}$ $sqrtx$ using fractional exponents:

$= 6 \int x^{\frac{1}{2}} d x$ $=6intx^(1/2)dx$

Now, integrate using the rule:

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C, n \neq - 1$ $intx^ndx=x^(n+1)/(n+1)+C" "" "," "" "n!=-1$

Giving:

$= 6 (\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}) + C = \frac{6 x^{\frac{3}{2}}}{\frac{3}{2}} + C = \frac{2}{3} (6 x^{\frac{3}{2}}) + C$ $=6(x^(1/2+1)/(1/2+1))+C=(6x^(3/2))/(3/2)+C=2/3(6x^(3/2))+C$

$= 4 x^{\frac{3}{2}} + C$ $=4x^(3/2)+C$

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What is the antiderivative of $6 \sqrt{x}$ $6sqrtx$ ?

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