What is the integral from-1 to 2 of [x] ?

The integral from-1 to 2 of [x] is 3/2

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integral from-1 to 2 of [x]

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

\frac{3}{2}

Decimal

1.5

Solution steps

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

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{- 1}^{2}

= [\frac{x ^{2}}{2}]_{- 1}^{2}

Compute the boundaries:

\frac{3}{2}

= \frac{3}{2}

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

\int_{0}^{2 π} 3 cos (t) d t

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

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

\int_{0}^{π} y sin (x) d x

What is the integral from-1 to 2 of [x] ?
The integral from-1 to 2 of [x] is 3/2