Answer
a) $-\sqrt {2}$
b) $\sqrt {2}$
c) $-\sqrt {2}$
d) 1
Work Step by Step
a) $\int^{-3}_{0}g(t)dt=-\int^{0}_{-3}g(t)dt=-\sqrt {2}$
b) $\int^{0}_{-3}g(u)du=\int^{0}_{-3}g(t)dt=\sqrt {2}$
c) $\int^{0}_{-3}[-g(x)]dx $
$=-\int^{0}_{-3}g(x)dx=-\int^{0}_{-3}g(t)dt=-\sqrt {2}$
d) $\int^{0}_{-3}\frac{g(r)}{\sqrt {2}}dr=\frac{1}{\sqrt 2}\int^{0}_{-3}g(t)dt $
$=\frac{1}{\sqrt 2}\times\sqrt 2=1$
