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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Answer

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

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