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

The integral from 1 to 3 of (2x+1)(3-x) is 26/3

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

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

\frac{26}{3}

Decimal

8.66666 \dots

Solution steps

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

Expand

(2 x + 1) (3 - x) : - 2 x^{2} + 5 x + 3

= \int_{1}^{3} - 2 x^{2} + 5 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_{1}^{3} 2 x^{2} d x + \int_{1}^{3} 5 x d x + \int_{1}^{3} 3 d x

\int_{1}^{3} 2 x^{2} d x = \frac{52}{3}

\int_{1}^{3} 5 x d x = 20

\int_{1}^{3} 3 d x = 6

= - \frac{52}{3} + 20 + 6

Simplify

- \frac{52}{3} + 20 + 6 : \frac{26}{3}

= \frac{26}{3}

What is the integral from 1 to 3 of (2x+1)(3-x) ?
The integral from 1 to 3 of (2x+1)(3-x) is 26/3