What is the integral from-1 to 4 of (-x^2+6x+8-x^2) ?

The integral from-1 to 4 of (-x^2+6x+8-x^2) is 125/3

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integral from-1 to 4 of (-x^2+6x+8-x^2)

\int_{- 1}^{4} (- x^{2} + 6 x + 8 - x^{2}) d x

\frac{125}{3}

Decimal

41.66666 \dots

Solution steps

\int_{- 1}^{4} - x^{2} + 6 x + 8 - x^{2} d x

Simplify

= \int_{- 1}^{4} - 2 x^{2} + 6 x + 8 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_{- 1}^{4} 2 x^{2} d x + \int_{- 1}^{4} 6 x d x + \int_{- 1}^{4} 8 d x

\int_{- 1}^{4} 2 x^{2} d x = \frac{130}{3}

\int_{- 1}^{4} 6 x d x = 45

\int_{- 1}^{4} 8 d x = 40

= - \frac{130}{3} + 45 + 40

Simplify

- \frac{130}{3} + 45 + 40 : \frac{125}{3}

= \frac{125}{3}

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What is the integral from-1 to 4 of (-x^2+6x+8-x^2) ?
The integral from-1 to 4 of (-x^2+6x+8-x^2) is 125/3