What is the integral from 1/2 to 4 of x^3-6x^2+9x+1 ?

The integral from 1/2 to 4 of x^3-6x^2+9x+1 is 679/64

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integral from 1/2 to 4 of x^3-6x^2+9x+1

\int_{\frac{1}{2}}^{4} x^{3} - 6 x^{2} + 9 x + 1 d x

\frac{679}{64}

Decimal

10.609375

Solution steps

\int_{\frac{1}{2}}^{4} x^{3} - 6 x^{2} + 9 x + 1 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_{\frac{1}{2}}^{4} x^{3} d x - \int_{\frac{1}{2}}^{4} 6 x^{2} d x + \int_{\frac{1}{2}}^{4} 9 x d x + \int_{\frac{1}{2}}^{4} 1 d x

\int_{\frac{1}{2}}^{4} x^{3} d x = \frac{4095}{64}

\int_{\frac{1}{2}}^{4} 6 x^{2} d x = \frac{511}{4}

\int_{\frac{1}{2}}^{4} 9 x d x = \frac{567}{8}

\int_{\frac{1}{2}}^{4} 1 d x = \frac{7}{2}

= \frac{4095}{64} - \frac{511}{4} + \frac{567}{8} + \frac{7}{2}

Simplify

\frac{4095}{64} - \frac{511}{4} + \frac{567}{8} + \frac{7}{2} : \frac{679}{64}

= \frac{679}{64}

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What is the integral from 1/2 to 4 of x^3-6x^2+9x+1 ?
The integral from 1/2 to 4 of x^3-6x^2+9x+1 is 679/64