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

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

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

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

\frac{13}{6}

Decimal

2.16666 \dots

Solution steps

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

\int x^{3} e^{x} d x

(x^{3} + y^{3}) d x + 3 x y^{2} d y = 0

\frac{d}{d x} ((8 x + 7)^{2})

\int \frac{1}{cos ( 3 x )} d x

- 9 x^{\frac{1}{2}} + x^{\frac{3}{2}}

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