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

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

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

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

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

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

Expand

x (2 x - sin (x^{2})) : 2 x^{2} - x sin (x^{2})

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

\int_{- 1}^{1} 2 x^{2} d x = \frac{4}{3}

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

= \frac{4}{3} - 0

Simplify

= \frac{4}{3}

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