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

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

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

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

\frac{16}{15}

Decimal Notation

1.06666 \dots

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

= \frac{16}{15}

\frac{\partial}{\partial x} (\frac{5}{x})

e^{2 x} \cdot 2

\int \frac{1}{1 + cosh ( x )} d x

f (x) = lo g_{3} (2 x^{2} + 4 x + 3)

\frac{\partial}{\partial y} (5 x^{3} y^{6})

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