What is the integral from 0 to 1 of [(x^2-1)-(-x+2)] ?

The integral from 0 to 1 of [(x^2-1)-(-x+2)] is -13/6

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integral from 0 to 1 of [(x^2-1)-(-x+2)]

\int_{0}^{1} [(x^{2} - 1) - (- x + 2)] d x

- \frac{13}{6}

Decimal

- 2.16666 \dots

Solution steps

\int_{0}^{1} x^{2} - 1 - (- x + 2) d x

Simplify

= \int_{0}^{1} x^{2} + x - 3 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} x^{2} d x + \int_{0}^{1} x d x - \int_{0}^{1} 3 d x

\int_{0}^{1} x^{2} d x = \frac{1}{3}

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

\int_{0}^{1} 3 d x = 3

= \frac{1}{3} + \frac{1}{2} - 3

Simplify

\frac{1}{3} + \frac{1}{2} - 3 : - \frac{13}{6}

= - \frac{13}{6}

\frac{\partial}{\partial x} (2 x - y)

\int 2 x (x^{2} + 2)^{5} d x

\frac{d}{d x} (\frac{5 x}{4})

tan (x), 0

y^{^{'}} - y = 1, y (0) = 0

What is the integral from 0 to 1 of [(x^2-1)-(-x+2)] ?
The integral from 0 to 1 of [(x^2-1)-(-x+2)] is -13/6