What is the integral from 1 to 4 of (sqrt(x)-x+2) ?

The integral from 1 to 4 of (sqrt(x)-x+2) is 19/6

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integral from 1 to 4 of (sqrt(x)-x+2)

\int_{1}^{4} (x - x + 2) d x

\frac{19}{6}

Decimal

3.16666 \dots

Solution steps

\int_{1}^{4} x - 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}^{4} x d x - \int_{1}^{4} x d x + \int_{1}^{4} 2 d x

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

\int_{1}^{4} x d x = \frac{15}{2}

\int_{1}^{4} 2 d x = 6

= \frac{14}{3} - \frac{15}{2} + 6

Simplify

\frac{14}{3} - \frac{15}{2} + 6 : \frac{19}{6}

= \frac{19}{6}

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What is the integral from 1 to 4 of (sqrt(x)-x+2) ?
The integral from 1 to 4 of (sqrt(x)-x+2) is 19/6