What is the integral from 1 to 16 of (x-7)/(sqrt(x)) ?

The integral from 1 to 16 of (x-7)/(sqrt(x)) is 0

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integral from 1 to 16 of (x-7)/(sqrt(x))

\int_{1}^{16} \frac{x - 7}{x} d x

0

Solution steps

\int_{1}^{16} \frac{x - 7}{x} d x

Expand

\frac{x - 7}{x} : x - \frac{7}{x}

= \int_{1}^{16} x - \frac{7}{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}^{16} x d x - \int_{1}^{16} \frac{7}{x} d x

\int_{1}^{16} x d x = 42

\int_{1}^{16} \frac{7}{x} d x = 42

= 42 - 42

Simplify

= 0

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What is the integral from 1 to 16 of (x-7)/(sqrt(x)) ?
The integral from 1 to 16 of (x-7)/(sqrt(x)) is 0