What is the integral from 1 to 3 of sin(r^2)r ?

The integral from 1 to 3 of sin(r^2)r is 1/2 (-cos(9)+cos(1))

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integral from 1 to 3 of sin(r^2)r

\int_{1}^{3} sin (r^{2}) r d r

\frac{1}{2} (- cos (9) + cos (1))

Decimal Notation

0.72571 \dots

Solution steps

\int_{1}^{3} sin (r^{2}) r d r

Apply u-substitution

: \int_{1}^{9} \frac{sin ( u )}{2} d u

= \int_{1}^{9} \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_{1}^{9} sin (u) d u

Use the common integral:

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

= \frac{1}{2} [- cos (u)]_{1}^{9}

Compute the boundaries:

- cos (9) + cos (1)

= \frac{1}{2} (- cos (9) + cos (1))

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What is the integral from 1 to 3 of sin(r^2)r ?
The integral from 1 to 3 of sin(r^2)r is 1/2 (-cos(9)+cos(1))