What is the integral from-1 to 0 of 2x^2(3x^3+1)^5 ?

The integral from-1 to 0 of 2x^2(3x^3+1)^5 is -7/3

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integral from-1 to 0 of 2x^2(3x^3+1)^5

\int_{- 1}^{0} 2 x^{2} (3 x^{3} + 1)^{5} d x

- \frac{7}{3}

Decimal

- 2.33333 \dots

Solution steps

\int_{- 1}^{0} 2 x^{2} (3 x^{3} + 1)^{5} d x

Take the constant out:

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

= 2 \cdot \int_{- 1}^{0} x^{2} (3 x^{3} + 1)^{5} d x

Expand

x^{2} (3 x^{3} + 1)^{5} : 243 x^{17} + 405 x^{14} + 270 x^{11} + 90 x^{8} + 15 x^{5} + x^{2}

= 2 \cdot \int_{- 1}^{0} 243 x^{17} + 405 x^{14} + 270 x^{11} + 90 x^{8} + 15 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 (\int_{- 1}^{0} 243 x^{17} d x + \int_{- 1}^{0} 405 x^{14} d x + \int_{- 1}^{0} 270 x^{11} d x + \int_{- 1}^{0} 90 x^{8} d x + \int_{- 1}^{0} 15 x^{5} d x + \int_{- 1}^{0} x^{2} d x)

\int_{- 1}^{0} 243 x^{17} d x = - \frac{27}{2}

\int_{- 1}^{0} 405 x^{14} d x = 27

\int_{- 1}^{0} 270 x^{11} d x = - \frac{45}{2}

\int_{- 1}^{0} 90 x^{8} d x = 10

\int_{- 1}^{0} 15 x^{5} d x = - \frac{5}{2}

\int_{- 1}^{0} x^{2} d x = \frac{1}{3}

= 2 (- \frac{27}{2} + 27 - \frac{45}{2} + 10 - \frac{5}{2} + \frac{1}{3})

Simplify

2 (- \frac{27}{2} + 27 - \frac{45}{2} + 10 - \frac{5}{2} + \frac{1}{3}) : - \frac{7}{3}

= - \frac{7}{3}

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What is the integral from-1 to 0 of 2x^2(3x^3+1)^5 ?
The integral from-1 to 0 of 2x^2(3x^3+1)^5 is -7/3