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

The integral from-1 to 2 of (4/3 x^3+4x^2+x) is 37/2

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

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

\frac{37}{2}

Decimal Notation

18.5

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{- 1}^{2} \frac{4}{3} x^{3} d x = 5

\int_{- 1}^{2} 4 x^{2} d x = 12

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

= 5 + 12 + \frac{3}{2}

Simplify

5 + 12 + \frac{3}{2} : \frac{37}{2}

= \frac{37}{2}

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What is the integral from-1 to 2 of (4/3 x^3+4x^2+x) ?
The integral from-1 to 2 of (4/3 x^3+4x^2+x) is 37/2