$int_{0}^{8}\frac{dx}{sqrt{1+sqrt3{x}}}$

Answer:

$4\sqrt{2} - 4$

Explanation:

Step1: Rewrite integrand

$\int_{0}^{8} x^{\frac{1}{3}} dx$

Step2: Apply power rule for integration

$\left[\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]{0}^{8} = \left[\frac{3}{4}x^{\frac{4}{3}}\right]{0}^{8}$

Step3: Evaluate at bounds

$\frac{3}{4}(8^{\frac{4}{3}}) - \frac{3}{4}(0^{\frac{4}{3}})$

Step4: Simplify the expression

$\frac{3}{4}(16) - 0 = 12$

Wait, correction: I misread the integrand. The correct integrand is $\sqrt[3]{x}=x^{1/3}$, but recalculating $8^{4/3}$: $8^{1/3}=2$, so $8^{4/3}=2^4=16$. Then $\frac{3}{4}*16=12$, 0 term is 0.

Answer:

$12$

Explanation:

Step1: Rewrite cube root as exponent

$\int_{0}^{8} x^{\frac{1}{3}} dx$

Step2: Use power rule for integrals

$\left[\frac{x^{\frac{4}{3}}}{\frac{4}{3}}\right]{0}^{8} = \left[\frac{3}{4}x^{\frac{4}{3}}\right]{0}^{8}$

Step3: Substitute upper limit $x=8$

$\frac{3}{4} \times (8^{\frac{4}{3}}) = \frac{3}{4} \times (2^4) = \frac{3}{4} \times 16$

Step4: Substitute lower limit $x=0$

$\frac{3}{4} \times 0^{\frac{4}{3}} = 0$

Step5: Compute final result

$12 - 0 = 12$