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}

\frac{d}{d x} (4 x (x^{2} - 4))

\int \frac{1}{2 x ^{2} - x} d x

f (x) = 5 x - 7

\frac{d}{d x} (x^{3} 2 x - 1)

\frac{d}{d x} (x^{3} - x^{2} - 2 x)

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