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

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

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

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

\frac{16}{15}

Decimal Notation

1.06666 \dots

Solution steps

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

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

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

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

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

Simplify

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

= \frac{16}{15}

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

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

\int_{0}^{2 π} 1 + cos (θ) d θ

\int_{- π}^{π} x cos^{2} (x) d x

\int_{0}^{1} x a x^{a - 1} d x

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