Calculate the integral: integral from 0 to 8(1+ sqrt(2x)+

Calculate the integral: $\int_0^8(1+\sqrt{2x}+\sqrt[3]{x}) \, dx$.

We will use the Newton-Leibniz formula by finding the antiderivative of each term separately.

Step 1: Rewrite the integrand

First, let's rewrite the terms using exponential notation:

\[\int_0^8(1+\sqrt{2x}+\sqrt[3]{x}) \, dx = \int_0^8\left(1+\sqrt{2}\cdot\sqrt{x}+x^{1/3}\right) dx\]

\[= \int_0^8\left(1+\sqrt{2}\cdot x^{1/2}+x^{1/3}\right) dx\]

Step 2: Find the antiderivative of each term

We integrate term by term.

Term 1: $\int 1 \, dx = x + C$

Term 2: $\int \sqrt{2} \cdot x^{1/2} \, dx = \sqrt{2} \int x^{1/2} \, dx$

Using the power rule $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$:

\[\sqrt{2} \int x^{1/2} \, dx = \sqrt{2} \cdot \frac{x^{1/2+1}}{1/2+1} = \sqrt{2} \cdot \frac{x^{3/2}}{3/2} = \sqrt{2} \cdot \frac{2x^{3/2}}{3} = \frac{2\sqrt{2}x^{3/2}}{3}\]

Term 3: $\int x^{1/3} \, dx = \frac{x^{1/3+1}}{1/3+1} = \frac{x^{4/3}}{4/3} = \frac{3x^{4/3}}{4}$

Step 3: Combine the antiderivatives

The antiderivative is:

\[F(x) = x + \frac{2\sqrt{2}x^{3/2}}{3} + \frac{3x^{4/3}}{4}\]

Step 4: Apply the Newton-Leibniz formula

\[\int_0^8(1+\sqrt{2x}+\sqrt[3]{x}) \, dx = \left[x + \frac{2\sqrt{2}x^{3/2}}{3} + \frac{3x^{4/3}}{4}\right]_0^8\]

Step 5: Evaluate at the upper limit $x = 8$

\[F(8) = 8 + \frac{2\sqrt{2} \cdot 8^{3/2}}{3} + \frac{3 \cdot 8^{4/3}}{4}\]

Calculate each term: the first term is $8$. For the second, $8^{3/2} = (\sqrt{8})^3 = (2\sqrt{2})^3 = 16\sqrt{2}$, so

\[\frac{2\sqrt{2} \cdot 16\sqrt{2}}{3} = \frac{32 \cdot 2}{3} = \frac{64}{3}\]

For the third, $8^{4/3} = (8^{1/3})^4 = 2^4 = 16$, so

\[\frac{3 \cdot 16}{4} = \frac{48}{4} = 12\]

Therefore:

\[F(8) = 8 + \frac{64}{3} + 12 = 20 + \frac{64}{3} = \frac{60 + 64}{3} = \frac{124}{3}\]

Step 6: Evaluate at the lower limit $x = 0$

\[F(0) = 0 + \frac{2\sqrt{2} \cdot 0^{3/2}}{3} + \frac{3 \cdot 0^{4/3}}{4} = 0\]

Step 7: Calculate the definite integral

\[\int_0^8(1+\sqrt{2x}+\sqrt[3]{x}) \, dx = F(8) - F(0) = \frac{124}{3} - 0 = \frac{124}{3}\]

Final answer:

\[\boxed{\int_0^8(1+\sqrt{2x}+\sqrt[3]{x}) \, dx = \frac{124}{3}}\]

In mixed number form: $\frac{124}{3} = 41\frac{1}{3}$ or approximately $41.333...$