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

The integral from-1 to 1 of (x^2+x^4sin(x)) is 2/3

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

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

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

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

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

\int_{- 1}^{1} x^{4} sin (x) d x = 0

= \frac{2}{3} + 0

Simplify

= \frac{2}{3}

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What is the integral from-1 to 1 of (x^2+x^4sin(x)) ?
The integral from-1 to 1 of (x^2+x^4sin(x)) is 2/3