What is the integral from 1 to-2 of (1/2) ?

The integral from 1 to-2 of (1/2) is -3/2

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integral from 1 to-2 of (1/2)

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

- \frac{3}{2}

Decimal Notation

- 1.5

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Integral of a constant:

\int a d x = a x

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

Compute the boundaries:

\frac{3}{2}

= - \frac{3}{2}

\frac{5}{s ^{2} + 5}

f (x) = 1 + 2 x - 3 x^{2}, at x = 0

\frac{d}{d x} (\frac{ln ( 1 - x ^{2} )}{x ^{2}})

\frac{d}{d x} (x^{5} + 1)

\int \frac{cot ( ln ( x ^{4} + 1 ) ) x ^{3}}{x ^{4} + 1} d x

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