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

The integral from-3 to 1 of (2x+1)(3+x)(1-x) is -32/3

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

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

- \frac{32}{3}

Decimal

- 10.66666 \dots

Solution steps

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

Multiply by the conjugate of

2 x + 1 : \frac{4 x ^{2} - 1}{2 x - 1}

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

Find undefined (singularity) points:

x = \frac{1}{2}

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

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

\int_{- 3}^{\frac{1}{2}} \frac{4 x ^{2} - 1}{2 x - 1} (3 + x) (1 - x) d x = - \frac{1127}{96}

\int_{\frac{1}{2}}^{1} \frac{4 x ^{2} - 1}{2 x - 1} (3 + x) (1 - x) d x = \frac{103}{96}

= - \frac{1127}{96} + \frac{103}{96}

Simplify

- \frac{1127}{96} + \frac{103}{96} : - \frac{32}{3}

= - \frac{32}{3}

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What is the integral from-3 to 1 of (2x+1)(3+x)(1-x) ?
The integral from-3 to 1 of (2x+1)(3+x)(1-x) is -32/3