Given $\int_{0}^{5} f(x) d x=17$ and $\int_{5}^{7} f(x) d x=10$, evaluate
(a) $\int_{0}^{7} f(x) d x$.
(b) $\int_{5}^{0} f(x) d x$.
(c) $\int_{5}^{5} f(x) d x$
(d) $\int_{0}^{5} 3 f(x) d x$
Answer
The integral from 0 to 5 of 3f(x) is 3 times the integral from 0 to 5 of f(x), because the integrand is multiplied by 3. So, \( \int_{0}^{5} 3 f(x) d x = 3 \times 17 = \boxed{51} \).
Step 1 ：The integral of a function over an interval [a, b] can be interpreted as the area under the curve of the function from a to b. The integral from 0 to 5 of f(x) is given as 17 and the integral from 5 to 7 of f(x) is given as 10.
Step 2 ：To find the integral from 0 to 7 of f(x), we can add the two given integrals together. So, \( \int_{0}^{7} f(x) d x = 17 + 10 = \boxed{27} \).
Step 3 ：The integral from 5 to 0 of f(x) is the negative of the integral from 0 to 5 of f(x), because the direction of integration is reversed. So, \( \int_{5}^{0} f(x) d x = -17 = \boxed{-17} \).
Step 4 ：The integral from 5 to 5 of f(x) is 0, because the limits of integration are the same, so there is no area under the curve. So, \( \int_{5}^{5} f(x) d x = \boxed{0} \).
Step 5 ：The integral from 0 to 5 of 3f(x) is 3 times the integral from 0 to 5 of f(x), because the integrand is multiplied by 3. So, \( \int_{0}^{5} 3 f(x) d x = 3 \times 17 = \boxed{51} \).
