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

Popular >

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}

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}

\int_{1}^{3} \frac{2 t + 3}{t + 1} d t

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

\int_{0}^{2} 2 π x 1 + 4 x^{2} d x

\int_{0}^{l n (x)} e^{- y} d y

\int_{0}^{\frac{x}{4}} tan^{3} (x) d x

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