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

The integral from-1 to 1 of 3x^2(2x^3+1)^2 is 14/3

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

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

\frac{14}{3}

Decimal

4.66666 \dots

Solution steps

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

Take the constant out:

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

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

Expand

x^{2} (2 x^{3} + 1)^{2} : 4 x^{8} + 4 x^{5} + x^{2}

= 3 \cdot \int_{- 1}^{1} 4 x^{8} + 4 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

= 3 (\int_{- 1}^{1} 4 x^{8} d x + \int_{- 1}^{1} 4 x^{5} d x + \int_{- 1}^{1} x^{2} d x)

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

\int_{- 1}^{1} 4 x^{5} d x = 0

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

= 3 (\frac{8}{9} + 0 + \frac{2}{3})

Simplify

3 (\frac{8}{9} + 0 + \frac{2}{3}) : \frac{14}{3}

= \frac{14}{3}

\int x^{3} - 3 x^{2} d x

x \to 2 + lim (\frac{2}{( x - 2 ) ^{5}})

\int 5 x e^{10 x} d x

x \to 0 + lim (\frac{x ^{2} - 3 x + 2}{x ^{3} - 2 x ^{2}})

ln (2 + 5 x)

What is the integral from-1 to 1 of 3x^2(2x^3+1)^2 ?
The integral from-1 to 1 of 3x^2(2x^3+1)^2 is 14/3