Question

calculate the definite integral.
int_{0}^{1} 7 sqrt7{x} d x
int_{0}^{1} 7 sqrt7{x} d x=square
(type an exact answer in simplified form.)

Explanation:

Step1: Rewrite the integrand

Rewrite \( \sqrt[7]{x}\) as \(x^{\frac{1}{7}}\). The integral becomes \(\int_{0}^{1}7x^{\frac{1}{7}}dx\).

Step2: Integrate using the power rule

The power rule for integration is \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)).
For \(\int7x^{\frac{1}{7}}dx\), since the integral of \(ax^{n}\) is \(a\frac{x^{n + 1}}{n + 1}+C\) (\(a = 7\), \(n=\frac{1}{7}\)), we have \(7\times\frac{x^{\frac{1}{7}+1}}{\frac{1}{7}+1}+C=7\times\frac{x^{\frac{8}{7}}}{\frac{8}{7}}+C=\frac{49}{8}x^{\frac{8}{7}}+C\).

Step3: Evaluate the definite integral

Using the fundamental theorem of calculus \(\int_{a}^{b}F^\prime(x)dx=F(b)-F(a)\).
Here \(F(x)=\frac{49}{8}x^{\frac{8}{7}}\), \(a = 0\), \(b = 1\).
\(F(1)-F(0)=\frac{49}{8}(1)^{\frac{8}{7}}-\frac{49}{8}(0)^{\frac{8}{7}}\).
Since \(1^{\frac{8}{7}}=1\) and \(0^{\frac{8}{7}}=0\), we get \(\frac{49}{8}-0\).

Answer:

\(\frac{49}{8}\)