What is the integral from-2 to 3 of (10x^3-4x)(x+2) ?

The integral from-2 to 3 of (10x^3-4x)(x+2) is 2425/3

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integral from-2 to 3 of (10x^3-4x)(x+2)

\int_{- 2}^{3} (10 x^{3} - 4 x) (x + 2) d x

\frac{2425}{3}

Decimal Notation

808.33333 \dots

Solution steps

\int_{- 2}^{3} (10 x^{3} - 4 x) (x + 2) d x

Multiply by the conjugate of

10 x^{3} - 4 x : \frac{100 x ^{6} - 16 x ^{2}}{10 x ^{3} + 4 x}

= \int_{- 2}^{3} \frac{100 x ^{6} - 16 x ^{2}}{10 x ^{3} + 4 x} (x + 2) d x

Find undefined (singularity) points:

x = 0

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_{- 2}^{0} \frac{100 x ^{6} - 16 x ^{2}}{10 x ^{3} + 4 x} (x + 2) d x + \int_{0}^{3} \frac{100 x ^{6} - 16 x ^{2}}{10 x ^{3} + 4 x} (x + 2) d x

\int_{- 2}^{0} \frac{100 x ^{6} - 16 x ^{2}}{10 x ^{3} + 4 x} (x + 2) d x = - \frac{32}{3}

\int_{0}^{3} \frac{100 x ^{6} - 16 x ^{2}}{10 x ^{3} + 4 x} (x + 2) d x = 819

= - \frac{32}{3} + 819

Simplify

- \frac{32}{3} + 819 : \frac{2425}{3}

= \frac{2425}{3}

f (x) = x sin (\frac{2}{x})

x \to 2 + lim (2 - x^{2})

- \frac{6}{x ^{2}} - \frac{5}{x}

x \to 2 lim (x^{2} - x - 2)

x \to - 2 lim (1)

What is the integral from-2 to 3 of (10x^3-4x)(x+2) ?
The integral from-2 to 3 of (10x^3-4x)(x+2) is 2425/3