What is the integral from 3 to 2 of (5x^2+4x+9) ?

The integral from 3 to 2 of (5x^2+4x+9) is -152/3

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integral from 3 to 2 of (5x^2+4x+9)

\int_{3}^{2} (5 x^{2} + 4 x + 9) d x

- \frac{152}{3}

Decimal Notation

- 50.66666 \dots

Solution steps

\int_{3}^{2} 5 x^{2} + 4 x + 9 d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{2}^{3} 5 x^{2} + 4 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_{2}^{3} 5 x^{2} d x + \int_{2}^{3} 4 x d x + \int_{2}^{3} 9 d x)

\int_{2}^{3} 5 x^{2} d x = \frac{95}{3}

\int_{2}^{3} 4 x d x = 10

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

= - (\frac{95}{3} + 10 + 9)

Simplify

- (\frac{95}{3} + 10 + 9) : - \frac{152}{3}

= - \frac{152}{3}

y^{^{''}} + 4 y = 5 sec (2 t)

y^{^{''}} (e^{x} + 1) + y^{^{'}} = 0

\int x^{3} \cdot 4 - x^{2} d x

\frac{4}{3} π r^{2}

t

What is the integral from 3 to 2 of (5x^2+4x+9) ?
The integral from 3 to 2 of (5x^2+4x+9) is -152/3