What is the integral from 1 to 4 of (4x^2-9x) ?

The integral from 1 to 4 of (4x^2-9x) is 33/2

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integral from 1 to 4 of (4x^2-9x)

\int_{1}^{4} (4 x^{2} - 9 x) d x

\frac{33}{2}

Decimal Notation

16.5

Solution steps

\int_{1}^{4} 4 x^{2} - 9 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}^{4} 4 x^{2} d x - \int_{1}^{4} 9 x d x

\int_{1}^{4} 4 x^{2} d x = 84

\int_{1}^{4} 9 x d x = \frac{135}{2}

= 84 - \frac{135}{2}

Simplify

84 - \frac{135}{2} : \frac{33}{2}

= \frac{33}{2}

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What is the integral from 1 to 4 of (4x^2-9x) ?
The integral from 1 to 4 of (4x^2-9x) is 33/2