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

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

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integral from 0 to 2 of sin(x^2)x

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

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

Decimal Notation

0.82682 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Compute the boundaries:

- cos (4) + 1

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

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