What is the integral from-1 to 3 of (3x-8)(x+3) ?

The integral from-1 to 3 of (3x-8)(x+3) is -64

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integral from-1 to 3 of (3x-8)(x+3)

\int_{- 1}^{3} (3 x - 8) (x + 3) d x

- 64

Solution steps

\int_{- 1}^{3} (3 x - 8) (x + 3) d x

Apply Integration By Parts

: [(3 x - 8) (\frac{x ^{2}}{2} + 3 x) - \int 3 (\frac{x ^{2}}{2} + 3 x) d x]_{- 1}^{3}

= [(3 x - 8) (\frac{x ^{2}}{2} + 3 x) - \int 3 (\frac{x ^{2}}{2} + 3 x) d x]_{- 1}^{3}

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

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

Compute the boundaries:

- 64

= - 64

\int_{0}^{4} 5 x^{2} d x

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

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

\int_{0}^{1 - x} 1 d y

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

What is the integral from-1 to 3 of (3x-8)(x+3) ?
The integral from-1 to 3 of (3x-8)(x+3) is -64