What is the integral from-1 to 1 of (x^4-x^6) ?

The integral from-1 to 1 of (x^4-x^6) is 4/35

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integral from-1 to 1 of (x^4-x^6)

\int_{- 1}^{1} (x^{4} - x^{6}) d x

\frac{4}{35}

Decimal

0.11428 \dots

Solution steps

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

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

\int_{- 1}^{1} x^{6} d x = \frac{2}{7}

= \frac{2}{5} - \frac{2}{7}

Simplify

\frac{2}{5} - \frac{2}{7} : \frac{4}{35}

= \frac{4}{35}

\int_{0}^{3} 5^{x} d x

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\int_{1}^{5} \frac{4 ( ln ( x ) ) ^{3}}{x} d x

\int_{0}^{t} sin (t - x) sin (x) d x

\int_{2}^{3} 8 x^{3} + 3 x^{2} + 6 x d x

What is the integral from-1 to 1 of (x^4-x^6) ?
The integral from-1 to 1 of (x^4-x^6) is 4/35