What is the integral from 0 to 1 of (2x^2+3x-4) ?

The integral from 0 to 1 of (2x^2+3x-4) is -11/6

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integral from 0 to 1 of (2x^2+3x-4)

\int_{0}^{1} (2 x^{2} + 3 x - 4) d x

- \frac{11}{6}

Decimal Notation

- 1.83333 \dots

Solution steps

\int_{0}^{1} 2 x^{2} + 3 x - 4 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}^{1} 2 x^{2} d x + \int_{0}^{1} 3 x d x - \int_{0}^{1} 4 d x

\int_{0}^{1} 2 x^{2} d x = \frac{2}{3}

\int_{0}^{1} 3 x d x = \frac{3}{2}

\int_{0}^{1} 4 d x = 4

= \frac{2}{3} + \frac{3}{2} - 4

Simplify

\frac{2}{3} + \frac{3}{2} - 4 : - \frac{11}{6}

= - \frac{11}{6}

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What is the integral from 0 to 1 of (2x^2+3x-4) ?
The integral from 0 to 1 of (2x^2+3x-4) is -11/6