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}

x^{^{''}} + x = 0, x (\frac{π}{2}) = 0, x^{^{'}} (\frac{π}{2}) = 1

\int a^{x} x d x

y = 2 x^{2}, y = 8 x - 2 x^{2}

\frac{\partial}{\partial y} (x^{2} + y^{2} - 13)

f (x) = - x^{2} - 7 x + 44, at x = 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