What is the integral from 0 to 3 of (x^2-5x+9) ?

The integral from 0 to 3 of (x^2-5x+9) is 27/2

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integral from 0 to 3 of (x^2-5x+9)

\int_{0}^{3} (x^{2} - 5 x + 9) d x

\frac{27}{2}

Decimal Notation

13.5

Solution steps

\int_{0}^{3} x^{2} - 5 x + 9 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_{0}^{3} x^{2} d x - \int_{0}^{3} 5 x d x + \int_{0}^{3} 9 d x

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

\int_{0}^{3} 5 x d x = \frac{45}{2}

\int_{0}^{3} 9 d x = 27

= 9 - \frac{45}{2} + 27

Simplify

9 - \frac{45}{2} + 27 : \frac{27}{2}

= \frac{27}{2}

\int_{2}^{4} x^{2} + 5 x + 7 + 1 d x

\int_{- 7}^{4} \frac{1}{2} x d x

\int_{3}^{6} (2 x + 9) d x

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

\int_{0}^{4} π (y^{2})^{2} d y

What is the integral from 0 to 3 of (x^2-5x+9) ?
The integral from 0 to 3 of (x^2-5x+9) is 27/2