What is the integral from 1 to 5 of (6-x)-(8-3x) ?

The integral from 1 to 5 of (6-x)-(8-3x) is 16

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integral from 1 to 5 of (6-x)-(8-3x)

\int_{1}^{5} (6 - x) - (8 - 3 x) d x

16

Solution steps

\int_{1}^{5} 6 - x - (8 - 3 x) d x

Simplify

= \int_{1}^{5} 2 x - 2 d x

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}^{5} 2 x d x - \int_{1}^{5} 2 d x

\int_{1}^{5} 2 x d x = 24

\int_{1}^{5} 2 d x = 8

= 24 - 8

Simplify

= 16

\frac{\partial}{\partial y} (x e^{- y})

y = x^{3} + e^{x}

y^{^{''}} - 4 y^{^{'}} + 5 y = 1 + 5 x

\int 9 x + 6 d x

\int \frac{x}{x ^{4} + 3 x ^{2} - 4} d x

What is the integral from 1 to 5 of (6-x)-(8-3x) ?
The integral from 1 to 5 of (6-x)-(8-3x) is 16