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

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

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

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

\frac{1}{2}

Solution steps

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

Eliminate Absolutes

= \int_{- 1}^{0} (- x + sin (x)) x^{2} d x + \int_{0}^{1} (x + sin (x)) x^{2} d x

\int_{- 1}^{0} (- x + sin (x)) x^{2} d x = - cos (1) - 2 sin (1) + \frac{9}{4}

\int_{0}^{1} (x + sin (x)) x^{2} d x = cos (1) + 2 sin (1) - \frac{7}{4}

= - cos (1) - 2 sin (1) + \frac{9}{4} + cos (1) + 2 sin (1) - \frac{7}{4}

Simplify

- cos (1) - 2 sin (1) + \frac{9}{4} + cos (1) + 2 sin (1) - \frac{7}{4} : \frac{1}{2}

= \frac{1}{2}

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