Evaluate the Integral

Problem

(∫_−1^0)(t(1/3)−t(2/3)*d(t))

Solution

1. Identify the integral as a definite integral of a polynomial-like expression involving powers of t

2. Apply the power rule for integration, which states (∫_^)(tn*d(t))=(t(n+1))/(n+1) for n≠−1

3. Find the antiderivative of the first term: (∫_^)(t(1/3)*d(t))=(t(4/3))/(4/3)=3/4*t(4/3)

4. Find the antiderivative of the second term: (∫_^)(t(2/3)*d(t))=(t(5/3))/(5/3)=3/5*t(5/3)

5. Combine the terms to form the general antiderivative F(t)=3/4*t(4/3)−3/5*t(5/3)

6. Evaluate the limits of integration by calculating F(0)−F*(−1)

7. Substitute the upper limit t=0 3/4*(0)(4/3)−3/5*(0)(5/3)=0

8. Substitute the lower limit t=−1 3/4*(−1)(4/3)−3/5*(−1)(5/3)

9. Simplify the powers of −1 (−1)(4/3)=((−1)4)(1/3)=1(1/3)=1 and (−1)(5/3)=((−1)5)(1/3)=(−1)(1/3)=−1

10. Calculate the lower limit value: 3/4*(1)−3/5*(−1)=3/4+3/5

11. Find a common denominator to add the fractions: 15/20+12/20=27/20

12. Subtract the lower limit value from the upper limit value: 0−27/20=−27/20

Final Answer

(∫_−1^0)(t(1/3)−t(2/3)*d(t))=−27/20

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