Suppose $f$ is continuous on $[3,13]$ and $\int_{3}^{13} f(t) d t=3$. Compute each of the following values
(a) $\int_{7}^{7} f(t) d t=$
(b) $\int_{13}^{3} f(t) d t=$
(c) $\int_{3}^{13} 6 f(t) d t=$
(d) $\int_{3}^{13} f(t)+6 d t=$
(e) $\int_{3}^{7} f(t) d t+\int_{7}^{13} f(t) d t=$
(f) The average value of $f(x)$ on $[3,13]=$
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$ \text{The average value of } f(x) \text{ on } [3,13] = \frac{1}{13-3} \times \int_{3}^{13} f(t) d t = \frac{1}{10} \times 3 = 0.3 $

Steps

Step 1 ：$ \int_{7}^{7} f(t) d t = 0 $ since the upper and lower limits are the same

Step 2 ：$ \int_{13}^{3} f(t) d t = -\int_{3}^{13} f(t) d t = -3 $

Step 3 ：$ \int_{3}^{13} 6 f(t) d t = 6 \times \int_{3}^{13} f(t) d t = 6 \times 3 = 18 $

Step 4 ：$ \int_{3}^{13} f(t)+6 d t = \int_{3}^{13} f(t) d t + \int_{3}^{13} 6 d t = 3 + 6 \times (13 - 3) = 63 $

Step 5 ：$ \int_{3}^{7} f(t) d t + \int_{7}^{13} f(t) d t = \int_{3}^{13} f(t) d t = 3 $

Step 6 ：$ \text{The average value of } f(x) \text{ on } [3,13] = \frac{1}{13-3} \times \int_{3}^{13} f(t) d t = \frac{1}{10} \times 3 = 0.3 $