What is the integral from 0 to 4 of (sqrt(y)-y+2) ?

The integral from 0 to 4 of (sqrt(y)-y+2) is 16/3

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integral from 0 to 4 of (sqrt(y)-y+2)

\int_{0}^{4} (y - y + 2) d y

\frac{16}{3}

Decimal Notation

5.33333 \dots

Solution steps

\int_{0}^{4} y - y + 2 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_{0}^{4} y d y - \int_{0}^{4} y d y + \int_{0}^{4} 2 d y

\int_{0}^{4} y d y = \frac{16}{3}

\int_{0}^{4} y d y = 8

\int_{0}^{4} 2 d y = 8

= \frac{16}{3} - 8 + 8

Simplify

= \frac{16}{3}

t \to - 2 lim (\frac{t ^{2} - 4}{t + 2})

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What is the integral from 0 to 4 of (sqrt(y)-y+2) ?
The integral from 0 to 4 of (sqrt(y)-y+2) is 16/3