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

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

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

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

\frac{16}{3}

Decimal Notation

5.33333 \dots

Solution steps

\int_{0}^{4} 4 x - 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_{0}^{4} 4 x d x - \int_{0}^{4} 2 x d x

\int_{0}^{4} 4 x d x = \frac{64}{3}

\int_{0}^{4} 2 x d x = 16

= \frac{64}{3} - 16

Simplify

\frac{64}{3} - 16 : \frac{16}{3}

= \frac{16}{3}

2 x^{2} + 5 x

x \to 0 lim (x^{3} sin (\frac{1}{x}))

f (x) = 2 sec (x)

f (x) = π^{9}

\frac{18}{( x - 6 ) ^{2}}

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