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

Popular >

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

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

\frac{1}{2}

Decimal Notation

0.5

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}

\int_{0}^{1} 7 x e^{x} d x

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

\int_{z^{2}}^{8 - z^{2}} z^{2} d z

\int_{0}^{\frac{π}{2}} (cos (x))^{4} d x

\int_{0}^{t} 12 e^{2 x} - 12 e^{x} d x

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