What is the integral from-1 to 8 of (10x-x^2)-(3x-8) ?

The integral from-1 to 8 of (10x-x^2)-(3x-8) is 243/2

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integral from-1 to 8 of (10x-x^2)-(3x-8)

\int_{- 1}^{8} (10 x - x^{2}) - (3 x - 8) d x

\frac{243}{2}

Decimal

121.5

Solution steps

\int_{- 1}^{8} 10 x - x^{2} - (3 x - 8) d x

Simplify

= \int_{- 1}^{8} - x^{2} + 7 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}^{8} x^{2} d x + \int_{- 1}^{8} 7 x d x + \int_{- 1}^{8} 8 d x

\int_{- 1}^{8} x^{2} d x = 171

\int_{- 1}^{8} 7 x d x = \frac{441}{2}

\int_{- 1}^{8} 8 d x = 72

= - 171 + \frac{441}{2} + 72

Simplify

- 171 + \frac{441}{2} + 72 : \frac{243}{2}

= \frac{243}{2}

What is the integral from-1 to 8 of (10x-x^2)-(3x-8) ?
The integral from-1 to 8 of (10x-x^2)-(3x-8) is 243/2