To find the exact value of the definite integral [tex]\(\int_{-2}^6 \frac{4}{\sqrt[3]{x^2}} \, dx\)[/tex], we need to follow these steps using the Fundamental Theorem of Calculus.
1. Rewrite the integrand:
[tex]\[\frac{4}{\sqrt[3]{x^2}} = 4 \cdot x^{-2/3}\][/tex]
2. Find the antiderivative:
To find the antiderivative of [tex]\(4x^{-2/3}\)[/tex], we apply the integration rule for power functions:
[tex]\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad \text{for } n \neq -1\][/tex]
Here, [tex]\(n = -\frac{2}{3}\)[/tex], so we get:
[tex]\[
\int 4x^{-2/3} \, dx = 4 \cdot \int x^{-2/3} \, dx = 4 \cdot \left(\frac{x^{(-2/3) + 1}}{(-2/3) + 1}\right) + C
= 4 \cdot \left(\frac{x^{1/3}}{1/3}\right) + C
= 4 \cdot 3x^{1/3} + C
= 12x^{1/3} + C
\][/tex]
3. Apply the limits:
Now we evaluate this antiderivative from [tex]\(-2\)[/tex] to [tex]\(6\)[/tex]:
[tex]\[
\left[ 12x^{1/3} \right]_{-2}^{6} = 12 \left(6^{1/3}\right) - 12 \left((-2)^{1/3}\right)
\][/tex]
4. Analyze the cube roots:
The cube root of [tex]\(6\)[/tex] is straightforward. However, the cube root of [tex]\(-2\)[/tex] can be tricky. For real numbers, [tex]\((-2)^{1/3}\)[/tex] is defined as the real cube root of [tex]\(-2\)[/tex], which is approximately [tex]\(-1.2599\)[/tex].
Thus, we need to carefully consider the real part in this specific case.
Now, let's contrast the evaluated values:
[tex]\[
12 \left(6^{1/3}\right) = 12 (1.817) \approx 21.804
\][/tex]
[tex]\[
12 \left((-2)^{1/3}\right) = 12 (-1.259) = -15.108
\][/tex]
5. Summarize the integral value:
Summing these values:
[tex]\[
\left[ 12x^{1/3} \right]_{-2}^{6} = 12 \cdot 6^{1/3} - 12 \cdot (-2^{1/3})
= 12 \left(6^{1/3}\right) + 12 \left(2^{1/3}\right)
= 12 \left( 6^{1/3} + 2^{1/3} \right)
\][/tex]
From the given options, the correct answer is:
[tex]\[
\boxed{12(\sqrt[3]{6}+\sqrt[3]{2})}
\][/tex]
