What is the integral from 0 to 16 of (sqrt(x)-3) ?

The integral from 0 to 16 of (sqrt(x)-3) is -16/3

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integral from 0 to 16 of (sqrt(x)-3)

\int_{0}^{16} (x - 3) d x

- \frac{16}{3}

Decimal

- 5.33333 \dots

Solution steps

\int_{0}^{16} x - 3 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_{0}^{16} x d x - \int_{0}^{16} 3 d x

\int_{0}^{16} x d x = \frac{128}{3}

\int_{0}^{16} 3 d x = 48

= \frac{128}{3} - 48

Simplify

\frac{128}{3} - 48 : - \frac{16}{3}

= - \frac{16}{3}

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What is the integral from 0 to 16 of (sqrt(x)-3) ?
The integral from 0 to 16 of (sqrt(x)-3) is -16/3