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

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

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

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

\frac{256}{3}

Decimal Notation

85.33333 \dots

Solution steps

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

\int 8 t d t

f^{^{'}} (x) = 3 e^{2 x^{2}}

y^{^{'}} = \frac{1 - 4 y ^{2}}{3 x y}

y = - 6 cos (- 2 x)

x \to 0 + lim (x^{\frac{2}{3}})

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