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 $[3,13]=$ Submit Answer

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