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

The integral from 1 to 4 of (3x+sqrt(x)) is 163/6

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integral from 1 to 4 of (3x+sqrt(x))

\int_{1}^{4} (3 x + x) d x

\frac{163}{6}

Solution steps

\int_{1}^{4} 3 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} 3 x d x + \int_{1}^{4} x d x

\int_{1}^{4} 3 x d x = \frac{45}{2}

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

= \frac{45}{2} + \frac{14}{3}

Simplify

\frac{45}{2} + \frac{14}{3} : \frac{163}{6}

= \frac{163}{6}

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What is the integral from 1 to 4 of (3x+sqrt(x)) ?
The integral from 1 to 4 of (3x+sqrt(x)) is 163/6