What is the integral from 0 to 1 of xsin(1-x^2) ?

The integral from 0 to 1 of xsin(1-x^2) is 1/2 (-cos(1)+1)

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integral from 0 to 1 of xsin(1-x^2)

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

\frac{1}{2} (- cos (1) + 1)

Decimal

0.22984 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{0} - \frac{sin ( u )}{2} d u

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \frac{1}{2} \cdot \int_{0}^{1} sin (u) d u)

Use the common integral:

\int sin (u) d u = - cos (u)

= - (- \frac{1}{2} [- cos (u)]_{0}^{1})

Simplify

= \frac{1}{2} [- cos (u)]_{0}^{1}

Compute the boundaries:

- cos (1) + 1

= \frac{1}{2} (- cos (1) + 1)

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What is the integral from 0 to 1 of xsin(1-x^2) ?
The integral from 0 to 1 of xsin(1-x^2) is 1/2 (-cos(1)+1)