Question

evaluate the integral \\(\int \sqrt{-2 + 7s}\mathrm{d}s\\).\\(\int \sqrt{-2 + 7s}\mathrm{d}s = \square\\)

Explanation:

Step1: Use substitution method

Let \( u = -2 + 7s \), then \( du = 7ds \), so \( ds=\frac{1}{7}du \).

Step2: Rewrite the integral

Substitute \( u \) and \( ds \) into the integral:

$$ \int\sqrt{-2 + 7s}ds=\int\sqrt{u}\cdot\frac{1}{7}du=\frac{1}{7}\int u^{\frac{1}{2}}du $$

Step3: Integrate using power rule

The power rule for integration is \( \int x^n dx=\frac{x^{n + 1}}{n+1}+C \) (\( n
eq - 1 \)). For \( n=\frac{1}{2} \):

$$ \frac{1}{7}\int u^{\frac{1}{2}}du=\frac{1}{7}\cdot\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C=\frac{1}{7}\cdot\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{2}{21}u^{\frac{3}{2}}+C $$

Step4: Substitute back \( u \)

Replace \( u=-2 + 7s \) into the result:

$$ \frac{2}{21}(-2 + 7s)^{\frac{3}{2}}+C $$

Answer:

\(\frac{2}{21}(-2 + 7s)^{\frac{3}{2}}+C\) (where \( C \) is the constant of integration)