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

The integral from-2 to 3 of (4x-3)(x-3) is 325/6

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

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

\frac{325}{6}

Decimal

54.16666 \dots

Solution steps

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

Apply Integration By Parts

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

\int 4 (\frac{x ^{2}}{2} - 3 x) d x = 4 (\frac{x ^{3}}{6} - \frac{3 x ^{2}}{2})

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

Simplify

[(4 x - 3) (\frac{x ^{2}}{2} - 3 x) - 4 (\frac{x ^{3}}{6} - \frac{3 x ^{2}}{2})]_{- 2}^{3} : [\frac{1}{6} (8 x^{3} - 45 x^{2} + 54 x)]_{- 2}^{3}

= [\frac{1}{6} (8 x^{3} - 45 x^{2} + 54 x)]_{- 2}^{3}

Compute the boundaries:

\frac{325}{6}

= \frac{325}{6}

What is the integral from-2 to 3 of (4x-3)(x-3) ?
The integral from-2 to 3 of (4x-3)(x-3) is 325/6