We are tasked with evaluating the integral:

$\int 20t^2 \sqrt[3]{t^{\frac{1}{3}}} \, dt$

Step 1: Simplify the integrand

First, let's simplify $\sqrt[3]{t^{\frac{1}{3}}}$. We know that $\sqrt[3]{x} = x^{1/3}$, so:

$\sqrt[3]{t^{\frac{1}{3}}} = t^{\frac{1}{9}}$

Thus, the integrand becomes:

$20t^2 \cdot t^{\frac{1}{9}} = 20t^{2 + \frac{1}{9}} = 20t^{\frac{19}{9}}$

Step 2: Apply the power rule for integration

The power rule states that:

$\int t^n \, dt = \frac{t^{n+1}}{n+1} + C$

In our case, $n = \frac{19}{9}$. Applying the power rule:

$\int 20t^{\frac{19}{9}} \, dt = 20 \cdot \frac{t^{\frac{19}{9} + 1}}{\frac{19}{9} + 1} + C$

Step 3: Simplify the exponent and the fraction

$\frac{19}{9} + 1 = \frac{19}{9} + \frac{9}{9} = \frac{28}{9}$

Thus, the integral becomes:

$20 \cdot \frac{t^{\frac{28}{9}}}{\frac{28}{9}} + C$

$= 20 \cdot \frac{9}{28} \cdot t^{\frac{28}{9}} + C$

$= \frac{180}{28} t^{\frac{28}{9}} + C$

Step 4: Simplify the constant

$\frac{180}{28} = \frac{90}{14} = \frac{45}{7}$

Thus, the final answer is:

$\frac{45}{7} t^{\frac{28}{9}} + C$

Final Answer:

$\int 20t^2 \sqrt[3]{t^{\frac{1}{3}}} \, dt = \frac{45}{7} t^{\frac{28}{9}} + C$

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Do you want any further details or have any questions?

Here are 5 related questions for further exploration:

1. How would you evaluate $\int t^2 \sqrt[4]{t^{\frac{1}{2}}} \, dt$?
2. Can you generalize this method to $\int t^n \sqrt[k]{t^m} \, dt$?
3. What happens if you differentiate the final result to verify the solution?
4. How would you solve $\int t^{\frac{5}{3}} \, dt$?
5. How does this integral change if the power of $t$ in the integrand is negative?

Tip: When integrating functions with fractional exponents, always simplify the terms before applying the power rule.