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

The integral from-1 to 1 of (3x^2-1)^2 is 8/5

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

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

\frac{8}{5}

Decimal

1.6

Solution steps

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

Expand

(3 x^{2} - 1)^{2} : 9 x^{4} - 6 x^{2} + 1

= \int_{- 1}^{1} 9 x^{4} - 6 x^{2} + 1 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}^{1} 9 x^{4} d x - \int_{- 1}^{1} 6 x^{2} d x + \int_{- 1}^{1} 1 d x

\int_{- 1}^{1} 9 x^{4} d x = \frac{18}{5}

\int_{- 1}^{1} 6 x^{2} d x = 4

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

= \frac{18}{5} - 4 + 2

Simplify

\frac{18}{5} - 4 + 2 : \frac{8}{5}

= \frac{8}{5}

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\int_{- \infty}^{0} e^{\frac{x}{2}} d x

What is the integral from-1 to 1 of (3x^2-1)^2 ?
The integral from-1 to 1 of (3x^2-1)^2 is 8/5