What is the integral from 2 to 4 of (3x-1+2x^2) ?

The integral from 2 to 4 of (3x-1+2x^2) is 160/3

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integral from 2 to 4 of (3x-1+2x^2)

\int_{2}^{4} (3 x - 1 + 2 x^{2}) d x

\frac{160}{3}

Decimal

53.33333 \dots

Solution steps

\int_{2}^{4} 3 x - 1 + 2 x^{2} 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}^{4} 3 x d x - \int_{2}^{4} 1 d x + \int_{2}^{4} 2 x^{2} d x

\int_{2}^{4} 3 x d x = 18

\int_{2}^{4} 1 d x = 2

\int_{2}^{4} 2 x^{2} d x = \frac{112}{3}

= 18 - 2 + \frac{112}{3}

Simplify

18 - 2 + \frac{112}{3} : \frac{160}{3}

= \frac{160}{3}

What is the integral from 2 to 4 of (3x-1+2x^2) ?
The integral from 2 to 4 of (3x-1+2x^2) is 160/3