What is the integral of (4x^3-2)2x^2 ?

The integral of (4x^3-2)2x^2 is 4((x^6)/3-(x^3)/3)+C

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integral of (4x^3-2)2x^2

\int (4 x^{3} - 2) 2 x^{2} d x

4 (\frac{x ^{6}}{3} - \frac{x ^{3}}{3}) + C

Solution steps

\int (4 x^{3} - 2) \cdot 2 x^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int (4 x^{3} - 2) x^{2} d x

(4 x^{3} - 2) x^{2} = 2 (2 x^{3} - 1) x^{2}

= 2 \cdot \int 2 (2 x^{3} - 1) x^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot 2 \cdot \int (2 x^{3} - 1) x^{2} d x

Expand

(2 x^{3} - 1) x^{2} : 2 x^{5} - x^{2}

= 2 \cdot 2 \cdot \int 2 x^{5} - 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

= 2 \cdot 2 (\int 2 x^{5} d x - \int x^{2} d x)

\int 2 x^{5} d x = \frac{x ^{6}}{3}

\int x^{2} d x = \frac{x ^{3}}{3}

= 2 \cdot 2 (\frac{x ^{6}}{3} - \frac{x ^{3}}{3})

Simplify

= 4 (\frac{x ^{6}}{3} - \frac{x ^{3}}{3})

Add a constant to the solution

= 4 (\frac{x ^{6}}{3} - \frac{x ^{3}}{3}) + C

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What is the integral of (4x^3-2)2x^2 ?
The integral of (4x^3-2)2x^2 is 4((x^6)/3-(x^3)/3)+C