# UNRESOLVED: Integrating Piece-wise Definite Integrals

UNRESOLVED

So given a function under the following conditions:

$$f(t) = \begin{cases} 3t && t\lt 3\\ 10 && 3\le t\le 6\\ -\frac{1}{3}t + 6 && 6\le t \le 9 \end{cases}$$

How would I go about finding the area of $$\int_0^9 f(t) \ dt$$

My logic has been to iterate each piece for the given condition i.e. Evaluate $3t$ from the $t$ value $0$ to $3$ et. cetera.

So for $\ \ 0-3 = 18,\\ \ \ 3-6 = 40,\\ \ \ 6-9 = 14$

Therefore $\int_0^9 f(t) \ dt = 72$?

Another idea of mine, simplify the formulae for each condition which is $\frac{8(t+6)}{3}$ and then integrate $f(t)$ as $$\int_0^9\frac{8(t+6)}{3}dt=252$$

• Your logic is correct, however you have computed the integrals incorrectly – 123 Jun 5 '17 at 5:28
• What is it that I have done incorrectly? – Computer Jun 5 '17 at 5:31
• Your function must be given as $3\le t < 6$. You know why? – farruhota Jun 11 '17 at 2:33