What is the integral from-1 to 7 of (2x+7)-(x^2-4x) ?

The integral from-1 to 7 of (2x+7)-(x^2-4x) is 256/3

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integral from-1 to 7 of (2x+7)-(x^2-4x)

\int_{- 1}^{7} (2 x + 7) - (x^{2} - 4 x) d x

\frac{256}{3}

Decimal Notation

85.33333 \dots

Solution steps

\int_{- 1}^{7} 2 x + 7 - (x^{2} - 4 x) d x

Simplify

= \int_{- 1}^{7} - x^{2} + 6 x + 7 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}^{7} x^{2} d x + \int_{- 1}^{7} 6 x d x + \int_{- 1}^{7} 7 d x

\int_{- 1}^{7} x^{2} d x = \frac{344}{3}

\int_{- 1}^{7} 6 x d x = 144

\int_{- 1}^{7} 7 d x = 56

= - \frac{344}{3} + 144 + 56

Simplify

- \frac{344}{3} + 144 + 56 : \frac{256}{3}

= \frac{256}{3}

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What is the integral from-1 to 7 of (2x+7)-(x^2-4x) ?
The integral from-1 to 7 of (2x+7)-(x^2-4x) is 256/3