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

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

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

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

Apply Integration By Parts

: [(3 x - 4) (\frac{x ^{2}}{2} - x) - \int 3 (\frac{x ^{2}}{2} - x) d x]_{1}^{2}

= [(3 x - 4) (\frac{x ^{2}}{2} - x) - \int 3 (\frac{x ^{2}}{2} - x) d x]_{1}^{2}

\int 3 (\frac{x ^{2}}{2} - x) d x = 3 (\frac{x ^{3}}{6} - \frac{x ^{2}}{2})

= [(3 x - 4) (\frac{x ^{2}}{2} - x) - 3 (\frac{x ^{3}}{6} - \frac{x ^{2}}{2})]_{1}^{2}

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

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