What is the integral from 4 to 16 of (sqrt(y)-y/4) ?

The integral from 4 to 16 of (sqrt(y)-y/4) is 22/3

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integral from 4 to 16 of (sqrt(y)-y/4)

\int_{4}^{16} (y - \frac{y}{4}) d y

\frac{22}{3}

Solution steps

\int_{4}^{16} y - \frac{y}{4} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{4}^{16} y d y - \int_{4}^{16} \frac{y}{4} d y

\int_{4}^{16} y d y = \frac{112}{3}

\int_{4}^{16} \frac{y}{4} d y = 30

= \frac{112}{3} - 30

Simplify

\frac{112}{3} - 30 : \frac{22}{3}

= \frac{22}{3}

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

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

\int_{0}^{4} 15 - 2 x - 7 d x

\int_{1}^{4} \frac{1 + t}{t} d t

\int_{0}^{5} x e^{x - 5} d x

What is the integral from 4 to 16 of (sqrt(y)-y/4) ?
The integral from 4 to 16 of (sqrt(y)-y/4) is 22/3