How can to get $v(t)$ out of this acceleration graph? [closed]

How can I get v(t) out of this acceleration graph?

• Area under the graph. Note it’s zero for the middle interval, and negative in the third. – joseph h Mar 19 at 6:07

$$a(t)=\frac{\mathrm{d}v(t)}{\mathrm{d}t}$$ $$\implies v(t)=\int a(t)\,\mathrm{d}t+C$$ If you would like to use explicit limits of integration, consider the particle at two times $$t_1$$ and $$t_2$$. Then, we can separate and integrate both sides: $$\int_{t_1}^{t_2}a(t)\,\mathrm{d}t=\int_{v(t_1)}^{v(t_2)}\,\mathrm{d}v(t)=v(t_2)-v(t_1)$$
• To start, use $a(t)=1$ for $0<t<3$. – G. Smith Mar 19 at 6:18