Question

$\int\limits^2_{-3} \, dx / \ sqrt{7-3x}$

Answer 1

$\int _{-3}^2\, \frac{dx}{\sqrt{7-3x}}=[\, d(7-3x)=-3\, dx\, ]=-\frac{1} {3}\int_{-3}^2\frac{d(7-3x)}{\sqrt{7-3x}}=\\\\=-\frac{1}{3}\cdot 2\sqrt{7-3x}\, |_{-3}^2=-\frac{2}{3}(\sqrt{7-6}-\sqrt{7+9})=-\frac{2}{3}(1-4)=2$