Question

) $I = \frac{x^4}{3\sqrt{x}}$

Answer 1

$\frac{2}{27} x^{9/2} + C$

Answer 2

Let the given integral be $I$.

$I = \int \frac{x^4}{3\sqrt{x}} dx$

First, simplify the integrand.

$\frac{x^4}{3\sqrt{x}} = \frac{x^4}{3x^{1/2}}$

Using the rule $a^m / a^n = a^{m-n}$, we get:

$\frac{x^4}{3x^{1/2}} = \frac{1}{3} x^{4 - 1/2} = \frac{1}{3} x^{8/2 - 1/2} = \frac{1}{3} x^{7/2}$

Now, the integral becomes:

$I = \int \frac{1}{3} x^{7/2} dx$

We can pull the constant $\frac{1}{3}$ out of the integral:

$I = \frac{1}{3} \int x^{7/2} dx$

Using the power rule for integration, $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (where $n \neq -1$), with $n = \frac{7}{2}$:

$\int x^{7/2} dx = \frac{x^{7/2 + 1}}{7/2 + 1} + C$

$\int x^{7/2} dx = \frac{x^{9/2}}{9/2} + C$

Now, substitute this back into the expression for $I$:

$I = \frac{1}{3} \left( \frac{x^{9/2}}{9/2} \right) + C$

$I = \frac{1}{3} \cdot \frac{2}{9} x^{9/2} + C$

$I = \frac{2}{27} x^{9/2} + C$

Explanation of the solution:

The integral is $\int \frac{x^4}{3\sqrt{x}} dx$.

Rewrite the integrand as $\frac{1}{3} \frac{x^4}{x^{1/2}} = \frac{1}{3} x^{4 - 1/2} = \frac{1}{3} x^{7/2}$.

The integral becomes $\int \frac{1}{3} x^{7/2} dx$.

Pull the constant out: $\frac{1}{3} \int x^{7/2} dx$.

Apply the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ with $n = 7/2$.

$\int x^{7/2} dx = \frac{x^{7/2+1}}{7/2+1} + C = \frac{x^{9/2}}{9/2} + C$.

Multiply by the constant $\frac{1}{3}$: $\frac{1}{3} \left(\frac{x^{9/2}}{9/2}\right) + C = \frac{1}{3} \cdot \frac{2}{9} x^{9/2} + C = \frac{2}{27} x^{9/2} + C$.