What is the integral from 0 to 1 of (4x-4sqrt(x)) ?

The integral from 0 to 1 of (4x-4sqrt(x)) is -2/3

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integral from 0 to 1 of (4x-4sqrt(x))

\int_{0}^{1} (4 x - 4 x) d x

- \frac{2}{3}

Decimal Notation

- 0.66666 \dots

Solution steps

\int_{0}^{1} 4 x - 4 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}^{1} 4 x d x - \int_{0}^{1} 4 x d x

\int_{0}^{1} 4 x d x = 2

\int_{0}^{1} 4 x d x = \frac{8}{3}

= 2 - \frac{8}{3}

Simplify

2 - \frac{8}{3} : - \frac{2}{3}

= - \frac{2}{3}

\frac{d y}{d x} + 12 y = 8

\frac{d}{d x} (\frac{sin ( x ^{2} )}{ln ( 2 x + 8 )})

(1 + x)^{3}

\frac{d}{d x} (sec (\frac{π}{2}))

What is the integral from 0 to 1 of (4x-4sqrt(x)) ?
The integral from 0 to 1 of (4x-4sqrt(x)) is -2/3