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Using the Fundamental Theorem of Calculus, find the exact value of

[tex]\[ \int_{-2}^6 \frac{4}{\sqrt[3]{x^2}} \, dx \][/tex]

A. [tex]\( 12(\sqrt[3]{36}+\sqrt[3]{4}) \)[/tex]
B. [tex]\( 12(\sqrt[3]{36}-\sqrt[3]{4}) \)[/tex]
C. [tex]\( 6(\sqrt[3]{6}-\sqrt[3]{2}) \)[/tex]
D. [tex]\( 6(\sqrt[3]{6}+\sqrt[3]{2}) \)[/tex]
E. [tex]\( 12(\sqrt[3]{6}-\sqrt[3]{2}) \)[/tex]
F. [tex]\( 12(\sqrt[3]{6}+\sqrt[3]{2}) \)[/tex]



Answer :

GinnyAnswer
To find the exact value of the integral [tex]\(\int_{-2}^6 \frac{4}{\sqrt[3]{x^2}} \, dx\)[/tex], we can follow these steps:

### Step 1: Simplify the Integrand
First, simplify the integrand [tex]\(\frac{4}{\sqrt[3]{x^2}}\)[/tex]:
[tex]\[ \frac{4}{\sqrt[3]{x^2}} = 4 \cdot x^{-2/3} \][/tex]

### Step 2: Integrate the Function
Next, integrate [tex]\(4x^{-2/3}\)[/tex] using the power rule for integration:
[tex]\[ \int 4x^{-2/3} \, dx = 4 \int x^{-2/3} \, dx = 4 \left( \frac{x^{1 - 2/3}}{1 - 2/3} \right) = 4 \left( \frac{x^{1/3}}{1/3} \right) = 4 \cdot 3 x^{1/3} = 12 x^{1/3} \][/tex]

### Step 3: Apply the Fundamental Theorem of Calculus
Evaluate this antiderivative at the bounds [tex]\(-2\)[/tex] and [tex]\(6\)[/tex]:
[tex]\[ \int_{-2}^6 4 x^{-2/3} \, dx = \left. 12 x^{1/3} \right|_{-2}^6 = 12 \left( 6^{1/3} - (-2)^{1/3} \right) \][/tex]

### Step 4: Simplify the Evaluated Result
Simplify the expression:
[tex]\[ 12 \left( 6^{1/3} - (-2)^{1/3} \right) \][/tex]

Here, [tex]\((-2)^{1/3}\)[/tex] can be expressed as [tex]\(- (2)^{1/3}\)[/tex], so:
[tex]\[ 12 \left( 6^{1/3} - (-2)^{1/3} \right) = 12 (6^{1/3} + 2^{1/3}) \][/tex]

### Step 5: Match with Given Options
The exact value of the integral matches one of the given options:
[tex]\[ \boxed{12 (\sqrt[3]{6} + \sqrt[3]{2})} \][/tex]

Thus, the correct answer is:

[tex]\[ 12 (\sqrt[3]{6} + \sqrt[3]{2}) \][/tex]

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