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

The integral from-1 to 1 of (x^2-1)^2 is 16/15

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

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

\frac{16}{15}

Decimal Notation

1.06666 \dots

Solution steps

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

Expand

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

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

\int_{- 1}^{1} x^{4} d x = \frac{2}{5}

\int_{- 1}^{1} 2 x^{2} d x = \frac{4}{3}

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

= \frac{2}{5} - \frac{4}{3} + 2

Simplify

\frac{2}{5} - \frac{4}{3} + 2 : \frac{16}{15}

= \frac{16}{15}

\int \frac{x - 1}{x} d x

6 x y^{^{'}} + y = x^{4}

\frac{d}{d x} (e^{x} (3 x + 1)^{3})

\frac{d}{d x} (\frac{e ^{x}}{y})

f (x) = 3 x^{2} - ln (x)

What is the integral from-1 to 1 of (x^2-1)^2 ?
The integral from-1 to 1 of (x^2-1)^2 is 16/15