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

The integral from 0 to 1 of xsin(2x^2) is 1/4 (-cos(2)+1)

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

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

\frac{1}{4} (- cos (2) + 1)

Decimal Notation

0.35403 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{2} \frac{sin ( u )}{4} d u

= \int_{0}^{2} \frac{sin ( u )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{0}^{2} sin (u) d u

Use the common integral:

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

= \frac{1}{4} [- cos (u)]_{0}^{2}

Compute the boundaries:

- cos (2) + 1

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

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What is the integral from 0 to 1 of xsin(2x^2) ?
The integral from 0 to 1 of xsin(2x^2) is 1/4 (-cos(2)+1)