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

The integral from 0 to 1 of (3x^2+4x-1/2) is 5/2

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

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

\frac{5}{2}

Decimal Notation

2.5

Solution steps

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

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

\int_{0}^{1} 4 x d x = 2

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

= 1 + 2 - \frac{1}{2}

Simplify

1 + 2 - \frac{1}{2} : \frac{5}{2}

= \frac{5}{2}

\int_{- 1}^{1} - x^{2} + 9 d x

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

\int_{0}^{8} \frac{2 x - 4}{x + 1} d x

\int_{0}^{- 2.5} 100 x d x

What is the integral from 0 to 1 of (3x^2+4x-1/2) ?
The integral from 0 to 1 of (3x^2+4x-1/2) is 5/2