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

The integral from-1 to 2 of (x^2+x)(3x+1) is 99/4

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

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

\frac{99}{4}

Decimal Notation

24.75

Solution steps

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

Multiply by the conjugate of

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

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

Find undefined (singularity) points:

x = 1, 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_{- 1}^{0} \frac{x ^{4} - x ^{2}}{x ^{2} - x} (3 x + 1) d x + \int_{0}^{1} \frac{x ^{4} - x ^{2}}{x ^{2} - x} (3 x + 1) d x + \int_{1}^{2} \frac{x ^{4} - x ^{2}}{x ^{2} - x} (3 x + 1) d x

\int_{- 1}^{0} \frac{x ^{4} - x ^{2}}{x ^{2} - x} (3 x + 1) d x = \frac{1}{12}

\int_{0}^{1} \frac{x ^{4} - x ^{2}}{x ^{2} - x} (3 x + 1) d x = \frac{31}{12}

\int_{1}^{2} \frac{x ^{4} - x ^{2}}{x ^{2} - x} (3 x + 1) d x = \frac{265}{12}

= \frac{1}{12} + \frac{31}{12} + \frac{265}{12}

Simplify

\frac{1}{12} + \frac{31}{12} + \frac{265}{12} : \frac{99}{4}

= \frac{99}{4}

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What is the integral from-1 to 2 of (x^2+x)(3x+1) ?
The integral from-1 to 2 of (x^2+x)(3x+1) is 99/4