evaluate the definite integral.\n int_{0}^{1}x(6sqrt3{x}+5sqrt4{x})dx

Explanation:

Step1: Expand the integrand

First, expand $x(6\sqrt[3]{x}+5\sqrt[4]{x})$ to get $6x\cdot x^{\frac{1}{3}}+5x\cdot x^{\frac{1}{4}}=6x^{1 + \frac{1}{3}}+5x^{1+\frac{1}{4}}=6x^{\frac{4}{3}}+5x^{\frac{5}{4}}$.

Step2: Apply the power - rule for integration

The power - rule for integration is $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). So, $\int_{0}^{1}(6x^{\frac{4}{3}}+5x^{\frac{5}{4}})dx=6\int_{0}^{1}x^{\frac{4}{3}}dx+5\int_{0}^{1}x^{\frac{5}{4}}dx$. For $\int_{0}^{1}x^{\frac{4}{3}}dx=\left[\frac{x^{\frac{4}{3}+1}}{\frac{4}{3}+1}\right]{0}^{1}=\left[\frac{x^{\frac{7}{3}}}{\frac{7}{3}}\right]{0}^{1}=\frac{3}{7}(1 - 0)=\frac{3}{7}$. For $\int_{0}^{1}x^{\frac{5}{4}}dx=\left[\frac{x^{\frac{5}{4}+1}}{\frac{5}{4}+1}\right]{0}^{1}=\left[\frac{x^{\frac{9}{4}}}{\frac{9}{4}}\right]{0}^{1}=\frac{4}{9}(1 - 0)=\frac{4}{9}$.

Step3: Calculate the definite integral

$6\int_{0}^{1}x^{\frac{4}{3}}dx+5\int_{0}^{1}x^{\frac{5}{4}}dx=6\times\frac{3}{7}+5\times\frac{4}{9}=\frac{18}{7}+\frac{20}{9}$. Find a common denominator, which is $63$. Then $\frac{18}{7}+\frac{20}{9}=\frac{18\times9}{7\times9}+\frac{20\times7}{9\times7}=\frac{162}{63}+\frac{140}{63}=\frac{162 + 140}{63}=\frac{302}{63}$.

Answer:

$\frac{302}{63}$