What is the integral from 2 to 1 of [2f(s)+5g(s)] ?

The integral from 2 to 1 of [2f(s)+5g(s)] is -3f-(15g)/2

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integral from 2 to 1 of [2f(s)+5g(s)]

\int_{2}^{1} [2 f (s) + 5 g (s)] d s

- 3 f - \frac{15 g}{2}

Solution steps

\int_{2}^{1} 2 f s + 5 g s d s

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{2} 2 f s + 5 g s d s

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int_{1}^{2} 2 f s d s + \int_{1}^{2} 5 g s d s)

\int_{1}^{2} 2 f s d s = 3 f

\int_{1}^{2} 5 g s d s = \frac{15 g}{2}

= - (3 f + \frac{15 g}{2})

Simplify

= - 3 f - \frac{15 g}{2}

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What is the integral from 2 to 1 of [2f(s)+5g(s)] ?
The integral from 2 to 1 of [2f(s)+5g(s)] is -3f-(15g)/2