evaluate the definite integral.\n\\(\\int_{0}^{1}\\sqrt3{1 + 7x}dx\\)

Explanation:

Step1: Use substitution

Let $u = 1+7x$, then $du=7dx$. When $x = 0$, $u = 1$; when $x = 1$, $u=8$. And $dx=\frac{1}{7}du$.

Step2: Rewrite the integral

The integral $\int_{0}^{1}\sqrt[3]{1 + 7x}dx$ becomes $\frac{1}{7}\int_{1}^{8}u^{\frac{1}{3}}du$.

Step3: Integrate $u^{\frac{1}{3}}$

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int u^{\frac{1}{3}}du=\frac{u^{\frac{1}{3}+1}}{\frac{1}{3}+1}=\frac{3}{4}u^{\frac{4}{3}}+C$.

Step4: Evaluate the definite integral

$\frac{1}{7}\left[\frac{3}{4}u^{\frac{4}{3}}\right]{1}^{8}=\frac{3}{28}\left[u^{\frac{4}{3}}\right]{1}^{8}=\frac{3}{28}(8^{\frac{4}{3}}-1^{\frac{4}{3}})$. Since $8^{\frac{4}{3}}=(8^{\frac{1}{3}})^4 = 2^4=16$ and $1^{\frac{4}{3}} = 1$, we get $\frac{3}{28}(16 - 1)=\frac{3}{28}\times15=\frac{45}{28}$.

Answer:

$\frac{45}{28}$