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

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

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

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

\frac{8}{3}

Decimal Notation

2.66666 \dots

Solution steps

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

Simplify

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

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

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

= - \frac{16}{3} + 8

Simplify

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

= \frac{8}{3}

\frac{d}{d x} (1 - \frac{9}{x ^{2}})

x \to - 2 lim (\frac{x + 4}{x + 2})

\frac{d}{d x} ((- 16 x^{2} + 1)^{2})

\int_{2}^{5} \frac{5}{x ^{2} - 1} d x

x \to 5 - lim (\frac{x ^{5}}{( x - 5 )})

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