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

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

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integral from 0 to 4 of (sqrt(x)-(x-2))

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

\frac{16}{3}

Decimal Notation

5.33333 \dots

Solution steps

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

Simplify

= \int_{0}^{4} x - x + 2 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}^{4} x d x - \int_{0}^{4} x d x + \int_{0}^{4} 2 d x

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

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

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

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

Simplify

= \frac{16}{3}

y = \frac{- 10}{x ^{4}}

\frac{d}{d t} (2 t)

\frac{d y}{d x} = y^{2} e^{x} sin (3 x), y (0) = 1

f (x) = \frac{x ^{2} - 4}{x - 4}

\frac{\partial}{\partial x} (e^{(- x^{2} - y^{2})})

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