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

The integral from-4 to 4 of (2-sqrt(x)) is 32/3-i 16/3

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

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

\frac{32}{3} - i \frac{16}{3}

Solution steps

\int_{- 4}^{4} 2 - 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_{- 4}^{4} 2 d x - \int_{- 4}^{4} x d x

\int_{- 4}^{4} 2 d x = 16

\int_{- 4}^{4} x d x = \frac{16}{3} + i \frac{16}{3}

= 16 - (\frac{16}{3} + i \frac{16}{3})

Simplify

16 - (\frac{16}{3} + i \frac{16}{3}) : \frac{32}{3} - i \frac{16}{3}

= \frac{32}{3} - i \frac{16}{3}

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What is the integral from-4 to 4 of (2-sqrt(x)) ?
The integral from-4 to 4 of (2-sqrt(x)) is 32/3-i 16/3