What is the integral from 1 to 4 of (6x+sqrt(x)) ?

The integral from 1 to 4 of (6x+sqrt(x)) is 149/3

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integral from 1 to 4 of (6x+sqrt(x))

\int_{1}^{4} (6 x + x) d x

\frac{149}{3}

Solution steps

\int_{1}^{4} 6 x + x 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}^{4} 6 x d x + \int_{1}^{4} x d x

\int_{1}^{4} 6 x d x = 45

\int_{1}^{4} x d x = \frac{14}{3}

= 45 + \frac{14}{3}

Simplify

45 + \frac{14}{3} : \frac{149}{3}

= \frac{149}{3}

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What is the integral from 1 to 4 of (6x+sqrt(x)) ?
The integral from 1 to 4 of (6x+sqrt(x)) is 149/3