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

The integral from 1 to 2 of (1-3x+2x^2) is 7/6

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

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

\frac{7}{6}

Decimal Notation

1.16666 \dots

Solution steps

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

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

\int_{1}^{2} 3 x d x = \frac{9}{2}

\int_{1}^{2} 2 x^{2} d x = \frac{14}{3}

= 1 - \frac{9}{2} + \frac{14}{3}

Simplify

1 - \frac{9}{2} + \frac{14}{3} : \frac{7}{6}

= \frac{7}{6}

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

\int_{0}^{1} 8 x^{- \frac{1}{5}} d x

\int_{0}^{5} t d t

\int_{2}^{4} x^{2} - 2 d x

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

What is the integral from 1 to 2 of (1-3x+2x^2) ?
The integral from 1 to 2 of (1-3x+2x^2) is 7/6