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

The integral from-1 to 2 of (x^2+1)(x-3) is -51/4

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

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

- \frac{51}{4}

Decimal

- 12.75

Solution steps

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

Multiply by the conjugate of

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

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

Find undefined (singularity) points:

x = 1

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

\int_{- 1}^{1} \frac{x ^{4} - 1}{x ^{2} - 1} (x - 3) d x = - 8

\int_{1}^{2} \frac{x ^{4} - 1}{x ^{2} - 1} (x - 3) d x = - \frac{19}{4}

= - 8 - \frac{19}{4}

Simplify

- 8 - \frac{19}{4} : - \frac{51}{4}

= - \frac{51}{4}

\int_{0}^{2} (2 x - 9) (4 x^{2} + 8) d x

\int_{0}^{1} (2 x - x^{2} - x^{2}) d x

\int_{0}^{4} 6 x^{2} d x

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

\int_{0}^{x} 2 - x d x

What is the integral from-1 to 2 of (x^2+1)(x-3) ?
The integral from-1 to 2 of (x^2+1)(x-3) is -51/4