Average rate of change of $f(x)=\int_{1}^{x}\sqrt{t^{3}+2}\,dt$ on $[0,3]$

Expert intro: This problem uses the definition of average rate of change together with the Fundamental Theorem of Calculus. The key is to evaluate $f(3)$ and $f(0)$ from the given integral, then apply the secant-slope formula on the interval $[0,3]$.

Detailed walkthrough

Use the average rate of change formula on $[0,3]$:

$\frac{f(3)-f(0)}{3-0}$

Given

$f(x)=\int_{1}^{x}\sqrt{t^{3}+2}\,dt,$

we compute:

$f(3)=\int_{1}^{3}\sqrt{t^{3}+2}\,dt$

and

$f(0)=\int_{1}^{0}\sqrt{t^{3}+2}\,dt=-\int_{0}^{1}\sqrt{t^{3}+2}\,dt.$

So

$f(3)-f(0)=\int_{1}^{3}\sqrt{t^{3}+2}\,dt+\int_{0}^{1}\sqrt{t^{3}+2}\,dt=\int_{0}^{3}\sqrt{t^{3}+2}\,dt.$

Therefore the average rate of change is

$\frac{1}{3}\int_{0}^{3}\sqrt{t^{3}+2}\,dt.$

Approximating this value gives

$\boxed{1.696}$

so the correct choice is C.

💡 Pitfall guide

A common mistake is to think the lower limit $1$ forces $f(0)=0$. It does not. When $x<1$, the integral changes sign because the upper and lower limits are reversed. Another mistake is to use $f'(x)$ instead of the average rate of change over the whole interval.

🔄 Real-world variant

If the interval were $[a,b]$ instead of $[0,3]$, the same method would give

$\frac{f(b)-f(a)}{b-a}.$

For any function defined by $f(x)=\int_{c}^{x}g(t)\,dt$, this simplifies to

$\frac{1}{b-a}\int_{a}^{b}g(t)\,dt,$

which is the average value of $g$ on $[a,b]$.

🔍 Related terms

Fundamental Theorem of Calculus, average rate of change, definite integral