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

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

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

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

\frac{3}{2}

Decimal

1.5

Solution steps

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

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

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

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

= \frac{4}{3} - (- \frac{1}{2}) - \frac{1}{3}

Simplify

\frac{4}{3} - (- \frac{1}{2}) - \frac{1}{3} : \frac{3}{2}

= \frac{3}{2}

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