What is the integral from x to 1 of sin(y^2) ?

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

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integral from x to 1 of sin(y^2)

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

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

Solution steps

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

Use the nonelementary integral:

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

= [\frac{π}{2} § (\frac{2}{π} y)]_{x}^{1}

Compute the boundaries:

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

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

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

\int_{0}^{1} \frac{1}{2} x d x

\int_{0}^{4} 4 x^{3} + \frac{2}{x + 1} d x

\int_{- 4}^{- 3} - x^{2} - 3 x d x

\int_{0}^{x} \frac{x}{x ^{2} + 1} d x

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