Let $f$ be the function defined by $f(x)=\int\limits_0^x (2t^3-15t^2+36t)   dt$. On which of the following intervals is the graph of $y=f(x)$ concave up?

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We need to use the second derivative test.

$$f(x)=\int\limits_0^x (2t^3-15t^2+36t)   dt$$ $$f'(x) = 2x^3-15x^2+36x$$ $$f''(x)=6x^2-30x+36$$

Finding points of inflection:

$$0 =6x^2-30x+36$$ $$0 =x^2-5x+6$$ $$0=(x-3)(x-2)$$

Testing $x=2$ and $x=3$:

$$\begin{array}{|c|c|c|c|} \hline \text{Interval} & -\infty \lt x \lt 2 & 2 \lt x \lt 3 & 3 \lt x \lt \infty \\ \hline \text{ Test Value} & x=-4 & x=2.5 & x= 4 \\ \hline \text{ Sign of }f''(x) & f''(-4) \gt 0 & f''(2.5) \lt 0 & f'(4) \gt 0 \\ \hline \text{Conclusion} & \text{Concave up} & \text{Concave down} & \text{concave up} \\ \hline \end{array}$$