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

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

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

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

- \frac{8}{3}

Decimal Notation

- 2.66666 \dots

Solution steps

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

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