Given ∫25f(x)dx=17 and ∫25g(x)dx=−2, evaluate the following. (a)∫25[f(x)+g(x)]dx (b)∫25[g(x)−f(x)]dx (c)∫252g(x)dx (d)∫253f(x)dx

Step 1 According to the standard rules of definite integrals, we have: 1.∫ab[f(x)±g(x)]dx=∫abf(x)dx±∫abg(x)dx 2.∫abcf(x)dx=c∫abf(x)dx Step 2 (a) ∫25[f(x)+g(x)]dx=∫25f(x)dx+∫25g(x)dx=(17)+(−2)=15 (b) ∫25[g(x)−f(x)]dx=∫25g(x)dx−∫25f(x)dx=(−2)−(17)=−19 (c) ∫252g(x)dx=2∫25g(x)dx=2(−2)=−4 (d) ∫253f(x)dx=3∫25f(x)dx=3(17)=51

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Answered question

2021-06-05

Given $\int_{2}^{5} f (x) d x = 17$ and $\int_{2}^{5} g (x) d x = - 2$ , evaluate the following.
(a) $\int_{2}^{5} [f (x) + g (x)] d x$
(b) $\int_{2}^{5} [g (x) - f (x)] d x$
(c) $\int_{2}^{5} 2 g (x) d x$
(d) $\int_{2}^{5} 3 f (x) d x$

Answer & Explanation

lobeflepnoumni

Skilled2021-06-06Added 99 answers

Step 1
According to the standard rules of definite integrals, we have:
1.

\int_{a}^{b} [f (x) \pm g (x)] d x = \int_{a}^{b} f (x) d x \pm \int_{a}^{b} g (x) d x

\int_{a}^{b} c f (x) d x = c \int_{a}^{b} f (x) d x

Step 2
(a)

\int_{2}^{5} [f (x) + g (x)] d x = \int_{2}^{5} f (x) d x + \int_{2}^{5} g (x) d x = (17) + (- 2) = 15

(b)

\int_{2}^{5} [g (x) - f (x)] d x = \int_{2}^{5} g (x) d x - \int_{2}^{5} f (x) d x = (- 2) - (17) = - 19

(c)

\int_{2}^{5} 2 g (x) d x = 2 \int_{2}^{5} g (x) d x = 2 (- 2) = - 4

(d)

\int_{2}^{5} 3 f (x) d x = 3 \int_{2}^{5} f (x) d x = 3 (17) = 51

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