What is the integral from 1 to 6 of (x^3+x^2+3) ?

The integral from 1 to 6 of (x^3+x^2+3) is 4925/12

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integral from 1 to 6 of (x^3+x^2+3)

\int_{1}^{6} (x^{3} + x^{2} + 3) d x

\frac{4925}{12}

Decimal

410.41666 \dots

Solution steps

\int_{1}^{6} x^{3} + x^{2} + 3 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}^{6} x^{3} d x + \int_{1}^{6} x^{2} d x + \int_{1}^{6} 3 d x

\int_{1}^{6} x^{3} d x = \frac{1295}{4}

\int_{1}^{6} x^{2} d x = \frac{215}{3}

\int_{1}^{6} 3 d x = 15

= \frac{1295}{4} + \frac{215}{3} + 15

Simplify

\frac{1295}{4} + \frac{215}{3} + 15 : \frac{4925}{12}

= \frac{4925}{12}

What is the integral from 1 to 6 of (x^3+x^2+3) ?
The integral from 1 to 6 of (x^3+x^2+3) is 4925/12