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

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

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

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

y (- cos (y) + 1)

Solution steps

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

Apply u-substitution

: \int_{0}^{y} sin (u) y d u

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

Take the constant out:

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

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

Use the common integral:

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

= y [- cos (u)]_{0}^{y}

Compute the boundaries:

- cos (y) + 1

= y (- cos (y) + 1)

\int_{0}^{5} 9 e^{3 x} d x

\int_{0}^{4} \frac{p}{9 + p ^{2}} d p

\int_{0}^{4} 2 x^{2} + 5 d x

\int_{- 1}^{1} ∣ x + 2 ∣ d x

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