What is the integral from-1 to 1 of x^4-2x^2+1 ?

The integral from-1 to 1 of x^4-2x^2+1 is 16/15

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integral from-1 to 1 of x^4-2x^2+1

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

\frac{16}{15}

Decimal Notation

1.06666 \dots

Solution steps

\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{y}{y ^{2} - 1} d y

sin (cos (tan (x)))

\frac{\partial}{\partial y} (\frac{x}{y + 2 z})

n = 1 \sum \infty (- 0.75)^{n}

\int x (x - 5) d x

What is the integral from-1 to 1 of x^4-2x^2+1 ?
The integral from-1 to 1 of x^4-2x^2+1 is 16/15