#### To determine

**To compute:**

The area between the graphs for 0≤t≤2 and interpret your result in the context.

#### Answer

0.867 inches

It rained 0.867 inches more at one location than the other.

#### Explanation

**1) Concept:**

The area between the graphs for 0≤t≤2 is nothing but integration of that region

**2) Given:**

i) ft=0.73t3-2t2+t+0.6

ii) gt=0.17t2-0.5t+1.1

**3) Formula:**

∫xn dx=xn+1n+1+C

4) **Calculation:**

From the graph, observe that gt≥f(t) on the interval [0, 2].

Therefore, the area between the two curves is

∫02gt-ftdt

Put expression gt, ft in the above integral, and solve the above integral.

=∫02(0.17t2-0.5t+1.1)-(0.73t3-2t2+t+0.6)dt

Combine like terms, and simplify.

=∫02-0.73t3+2.17t2-1.5t+0.5dt

Simplify integral.

=-0.73t44+2.17t33+1.5t22-0.5t02

Substitute limits.

=-0.73244+2.17233+1.5222-0.52-0

Simplify.

≈0.867

∫02gt-ftdt ≈0.867

Therefore, it rained 0.867 inches more at one location than the other.

**Conclusion:**

It rained 0.867 inches more at one location than the other.