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

The integral from 0 to 2 of (4x-x^2-x^2) is 8/3

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

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

\frac{8}{3}

Decimal Notation

2.66666 \dots

Solution steps

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

Simplify

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

\int_{0}^{2} 4 x d x = 8

\int_{0}^{2} 2 x^{2} d x = \frac{16}{3}

= 8 - \frac{16}{3}

Simplify

8 - \frac{16}{3} : \frac{8}{3}

= \frac{8}{3}

\int_{4}^{\infty} \frac{3}{x ^{5}} d x

\int_{1}^{3} 2 x^{2} + 3 d x

\int_{0}^{4} \frac{x - 1}{x ^{2} + x - 2} d x

\int_{8}^{\infty} 2 x^{- 3} d x

\int_{1}^{2} \frac{1}{x + y} d x

What is the integral from 0 to 2 of (4x-x^2-x^2) ?
The integral from 0 to 2 of (4x-x^2-x^2) is 8/3