What is the integral from 0 to 3 of (x+1)(2x-3) ?

The integral from 0 to 3 of (x+1)(2x-3) is 9/2

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integral from 0 to 3 of (x+1)(2x-3)

\int_{0}^{3} (x + 1) (2 x - 3) d x

\frac{9}{2}

Decimal Notation

4.5

Solution steps

\int_{0}^{3} (x + 1) (2 x - 3) d x

Expand

(x + 1) (2 x - 3) : 2 x^{2} - x - 3

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

\int_{0}^{3} 2 x^{2} d x = 18

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

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

= 18 - \frac{9}{2} - 9

Simplify

18 - \frac{9}{2} - 9 : \frac{9}{2}

= \frac{9}{2}

\frac{\partial}{\partial x} (4 x^{3} y^{3} + 3)

\frac{d}{d n} (2^{n})

\frac{x - 1}{x + 1}

x y^{^{'}} = y + x^{2} sin (x), y (π) = 0

x \to 0 - lim (x^{2} + 4)

What is the integral from 0 to 3 of (x+1)(2x-3) ?
The integral from 0 to 3 of (x+1)(2x-3) is 9/2