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

The integral from 1 to 2 of (x^3+3x^2-4x) is 19/4

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

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

\frac{19}{4}

Decimal

4.75

Solution steps

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

\int_{1}^{2} x^{3} d x = \frac{15}{4}

\int_{1}^{2} 3 x^{2} d x = 7

\int_{1}^{2} 4 x d x = 6

= \frac{15}{4} + 7 - 6

Simplify

\frac{15}{4} + 7 - 6 : \frac{19}{4}

= \frac{19}{4}

What is the integral from 1 to 2 of (x^3+3x^2-4x) ?
The integral from 1 to 2 of (x^3+3x^2-4x) is 19/4