Answer
$G(1)=0$
$G'(1)=-1$
$G'(2)=2$
$G(x)=\frac{x^{3}}{3}-2x+\frac{5}{3}$
Work Step by Step
$$G(x)=\int_{1}^{x}(t^{2}-2)dt$$
$$G(1)=\int_{1}^{1}(t^{2}-2)dt=0$$
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$$G(x)=\int_{1}^{x}(t^{2}-2)dt$$
Using the $FTC II$ it follows:
$$G'(x)=x^{2}-2$$
$$G'(1)=1^{2}-2=-1$$
$$G'(2)=2^{2}-2=2$$
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$$G(x)=\int_{1}^{x}(t^{2}-2)dt$$
Using the $FTC I$ it follows:
$$G(x)=[\frac{t^{3}}{3}-2t]_{1}^{x}$$
$$G(x)=\frac{x^{3}}{3}-2x-(\frac{1^{3}}{3}-2\cdot 1)$$
$$G(x)=\frac{x^{3}}{3}-2x+\frac{5}{3}$$
