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

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

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

\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 = \frac{8}{3}

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

= \frac{8}{3} - 2

Simplify

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

= \frac{2}{3}

\int \frac{x ^{5}}{( 3 - x ^{6} ) ^{4}} d x

y^{^{''}} + 7 y^{^{'}} + 12 y = 0

n = 5 \sum \infty \frac{n !}{n ^{n}}

\int \frac{1}{- t ^{2} + 6 t - 8} d t

\frac{d}{d x} (0.006 x^{3} + 0.02 x^{2} + 0.5 x)

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