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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 $$\int_0^9\frac{8(t+6)}{3}dt=252$$

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  • $\begingroup$ Your logic is correct, however you have computed the integrals incorrectly $\endgroup$ – 123 Jun 5 '17 at 5:28
  • $\begingroup$ What is it that I have done incorrectly? $\endgroup$ – Computer Jun 5 '17 at 5:31
  • $\begingroup$ Your function must be given as $3\le t < 6$. You know why? $\endgroup$ – farruhota Jun 11 '17 at 2:33
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\begin{align} \int_0^9\!f(t)\ dt&=\int_0^3\!f(t)\ dt+\int_3^6\!f(t)\ dt+\int_6^9\!f(t)\ dt\\ &=\int_0^3\!3t\ dt+\int_3^6\!10\ dt+\int_6^9\!\left(-\frac{1}{3}t + 6\right)\ dt\\ &=\frac32t^2\big|_0^3+10t\big|_3^6-\frac16t^2\big|_6^9+6t\big|_6^9\\ &=\frac32\times9+30-\frac{9^2-6^2}{6}+18\\ &=42 \end{align}

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