What is the integral of 10.2+2.5x+0.1x^2 ?

The integral of 10.2+2.5x+0.1x^2 is 10.2x+1.25x^2+0.03333…x^3+C

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integral of 10.2+2.5x+0.1x^2

\int 10.2 + 2.5 x + 0.1 x^{2} d x

10.2 x + 1.25 x^{2} + 0.03333 \dots x^{3} + C

Solution steps

\int 10.2 + 2.5 x + 0.1 x^{2} 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 10.2 d x + \int 2.5 x d x + \int 0.1 x^{2} d x

\int 10.2 d x = 10.2 x

\int 2.5 x d x = 1.25 x^{2}

\int 0.1 x^{2} d x = 0.03333 \dots x^{3}

= 10.2 x + 1.25 x^{2} + 0.03333 \dots x^{3}

Add a constant to the solution

= 10.2 x + 1.25 x^{2} + 0.03333 \dots x^{3} + C

\int a + b x + x^{5} d x

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x^{- 4}

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What is the integral of 10.2+2.5x+0.1x^2 ?
The integral of 10.2+2.5x+0.1x^2 is 10.2x+1.25x^2+0.03333…x^3+C