Answer
$\int_{0}^{3}g(t)dt=\frac{3}{2}$
$\int_{3}^{5}g(t)dt=0$
Work Step by Step
$$\int_{0}^{3}g(t)dt=\int_{0}^{1}g(t)dt+\int_{1}^{3}g(t)dt$$
The integral $\int_{0}^{1}g(t)dt$ represents the area of the region bounded between $g$,$x$ axis for $x$ between $0$ and $1$ which is the area of the triangle with base $1$ unit and height $1$ unit so:
$$\int_{0}^{1}g(t)dt=-\frac{1}{2} \cdot 1 \cdot 1=-\frac{1}{2}$$
The integral $\int_{1}^{3}g(t)dt$ represents the area of the region bounded between $g$,$x$ axis for $x$ between $1$ and $3$ which is the area of the triangle with base $2$ unit and height $2$ unit so:
$$\int_{1}^{3}g(t)dt=\frac{1}{2} \cdot 2 \cdot 2=2$$
so:
$$\int_{0}^{1}g(t)dt+\int_{1}^{3}g(t)dt$$
$$=-\frac{1}{2}+2=\frac{3}{2}$$
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$$\int_{3}^{5}g(t)dt=\int_{3}^{4}g(t)dt+\int_{4}^{5}g(t)dt$$
The integral $\int_{3}^{4}g(t)dt$ represents the area of the region bounded between $g$,$x$ axis for $x$ between $3$ and $4$ which is the area of the triangle with base $1$ unit and height $2$ unit so:
$$\int_{3}^{4}g(t)dt=\frac{1}{2} \cdot 1 \cdot 2=1$$
The integral $\int_{4}^{5}g(t)dt$ represents the area of the region bounded between $g$,$x$ axis for $x$ between $4$ and $5$ which is the area of the triangle with base $1$ unit and height $2$ unit so:
$$\int_{4}^{5}g(t)dt=-\frac{1}{2} \cdot 1 \cdot 2=-1$$
so:
$$\int_{3}^{5}g(t)dt=\int_{3}^{4}g(t)dt+\int_{4}^{5}g(t)dt$$
$$\int_{3}^{5}g(t)dt=1-1$$
$$\int_{3}^{5}g(t)dt=0$$
