Math Problem Statement

20t2t3\differentialDt\int20t^2\sqrt[3]{t}\differentialD t

Solution

We are tasked with solving the integral:

20t2t3dt\int 20t^2 \sqrt[3]{t} \, dt

Step 1: Rewrite the integrand

First, express t3\sqrt[3]{t} as a fractional exponent. We have:

t3=t1/3\sqrt[3]{t} = t^{1/3}

Thus, the integrand becomes:

20t2t1/3dt\int 20t^2 t^{1/3} \, dt

Step 2: Simplify the powers of tt

Using the property of exponents, tatb=ta+bt^a t^b = t^{a+b}, we combine the powers of tt:

t2t1/3=t2+1/3=t7/3t^2 \cdot t^{1/3} = t^{2 + 1/3} = t^{7/3}

Thus, the integral is now:

20t7/3dt\int 20 t^{7/3} \, dt

Step 3: Apply the power rule of integration

The power rule states that:

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

Here, n=73n = \frac{7}{3}, so we apply the power rule:

20t7/3dt=20t7/3+17/3+1+C\int 20 t^{7/3} \, dt = 20 \cdot \frac{t^{7/3 + 1}}{7/3 + 1} + C

Step 4: Simplify the exponent and the denominator

We simplify 7/3+17/3 + 1:

7/3+1=7/3+3/3=10/37/3 + 1 = 7/3 + 3/3 = 10/3

Thus, the integral becomes:

20t10/310/3+C20 \cdot \frac{t^{10/3}}{10/3} + C

Step 5: Simplify the fraction

Dividing by 10/310/3 is equivalent to multiplying by 3/103/10:

20310t10/3=6t10/320 \cdot \frac{3}{10} \cdot t^{10/3} = 6 t^{10/3}

Final Answer:

20t2t3dt=6t10/3+C\int 20 t^2 \sqrt[3]{t} \, dt = 6 t^{10/3} + C

Would you like further details or explanations on any step?

Related Questions:

  1. How do you integrate other fractional powers of tt?
  2. What is the general power rule for integration?
  3. How do you handle integrals with radicals and exponents?
  4. Can integration by substitution be applied here?
  5. How does this integral relate to definite integrals?

Tip:

Always simplify the integrand as much as possible before applying integration rules, especially by combining like terms or converting roots to fractional exponents.

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Exponentiation
Power Rule for Integration

Formulas

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

Theorems

Power Rule of Integration

Suitable Grade Level

Undergraduate Calculus or Advanced High School Calculus