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

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

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

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

x^{2} (- cos (z x) + 1)

Solution steps

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

Take the constant out:

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

= x^{2} \cdot \int_{0}^{x z} sin (y) d y

Use the common integral:

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

= x^{2} [- cos (y)]_{0}^{x z}

Compute the boundaries:

- cos (z x) + 1

= x^{2} (- cos (z x) + 1)

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

\int_{0}^{t} \frac{1}{x + 10} d x

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

\int_{1}^{e^{2}} (ln (x))^{2} d x

\int_{0}^{1} \frac{1}{( 5 - 3 x ) ^{2}} d x

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