3.6 Q-2

Question Statement

Evaluate the integral: $\int_{1}^{1} \left(x^{1/3} + 1\right) , dx$

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Background and Explanation

This problem involves the integration of a function over an interval where both limits are the same. When the upper and lower limits of the integral are equal, the result of the integral is always zero. This is because the area under the curve between two identical points is zero.

To solve this integral, we still apply the basic rules of integration, but we will see that the limits play a crucial role in the result.

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Solution

1. Set Up the Integral:

   We begin by separating the integral into two parts: $\int_{1}^{1} \left(x^{1/3} + 1\right) , dx = \int_{1}^{1} x^{1/3} , dx + \int_{1}^{1} 1 , dx$

2. Integrate $x^{1/3}$:

   Using the power rule: $\int x^{1/3} , dx = \frac{x^{(1/3) + 1}}{(1/3) + 1} = \frac{x^{4/3}}{4/3} = \frac{3x^{4/3}}{4}$

   Now apply the limits from 1 to 1: $\left[ \frac{3x^{4/3}}{4} \right]_{1}^{1} = \frac{3(1^{4/3})}{4} - \frac{3(1^{4/3})}{4} = 0$

3. Integrate 1:

   The integral of 1 is straightforward: $\int 1 , dx = x$

   Now apply the limits from 1 to 1: $\left[ x \right]_{1}^{1} = 1 - 1 = 0$

4. Combine the Results:

   Adding both integrals together: $0 + 0 = 0$

Thus, the value of the integral is: $\boxed{0}$

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Key Formulas or Methods Used

• Power rule for integration:
  $\int x^n , dx = \frac{x^{n+1}}{n+1} + C$
• The result of an integral where the upper and lower limits are the same is always zero.

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Summary of Steps

1. Split the integral into two parts:
   $\int_{1}^{1} (x^{1/3} + 1) , dx = \int_{1}^{1} x^{1/3} , dx + \int_{1}^{1} 1 , dx$

2. Integrate $x^{1/3}$ and apply limits:
   $\int_{1}^{1} x^{1/3} , dx = 0$

3. Integrate 1 and apply limits:
   $\int_{1}^{1} 1 , dx = 0$

4. Add the results:
   $0 + 0 = 0$

Thus, the final result is $0$.