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

The integral from 1 to 3 of sin(z^2) is sqrt(pi/2)\S(sqrt(2/pi)*3)-sqrt(pi/2)\S(sqrt(2/pi)*1)

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

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

\frac{π}{2} § (\frac{2}{π} \cdot 3) - \frac{π}{2} § (\frac{2}{π} \cdot 1)

Solution steps

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

Use the nonelementary integral:

\int sin (x^{2}) d x = \frac{π}{2} § (\frac{2}{π} x)

= [\frac{π}{2} § (\frac{2}{π} z)]_{1}^{3}

Compute the boundaries:

\frac{π}{2} § (\frac{2}{π} \cdot 3) - \frac{π}{2} § (\frac{2}{π} \cdot 1)

= \frac{π}{2} § (\frac{2}{π} \cdot 3) - \frac{π}{2} § (\frac{2}{π} \cdot 1)

\int_{k}^{x} \frac{1}{t} d t

\int_{0}^{1} \frac{x - 1}{x} d x

\int_{0}^{1} 3 t^{2} sin (t^{3}) d t

\int_{1}^{e} \frac{( ln ( x ^{4} ) )}{x} d x

\int_{1}^{l n (8)} e^{y} (y - 1) d y

What is the integral from 1 to 3 of sin(z^2) ?
The integral from 1 to 3 of sin(z^2) is sqrt(pi/2)\S(sqrt(2/pi)*3)-sqrt(pi/2)\S(sqrt(2/pi)*1)