What is the integral from-3 to 0 of (x^2-4x+7) ?

The integral from-3 to 0 of (x^2-4x+7) is 48

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integral from-3 to 0 of (x^2-4x+7)

\int_{- 3}^{0} (x^{2} - 4 x + 7) d x

48

Solution steps

\int_{- 3}^{0} x^{2} - 4 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_{- 3}^{0} x^{2} d x - \int_{- 3}^{0} 4 x d x + \int_{- 3}^{0} 7 d x

\int_{- 3}^{0} x^{2} d x = 9

\int_{- 3}^{0} 4 x d x = - 18

\int_{- 3}^{0} 7 d x = 21

= 9 - (- 18) + 21

Simplify

= 48

\int (x + 3) 2 - x d x

x \to 4 - lim (5 (x) - 7)

arccos (1 - x^{2})

y = 2 x^{2} + 5 x

\int (x + \frac{1}{9 x}) d x

What is the integral from-3 to 0 of (x^2-4x+7) ?
The integral from-3 to 0 of (x^2-4x+7) is 48