What is the integral from 0 to sin(y) of (y-2y) ?

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

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integral from 0 to sin(y) of (y-2y)

\int_{0}^{s i n (y)} (y - 2 y) d y

- \frac{1}{2} sin^{2} (y)

Solution steps

\int_{0}^{s i n (y)} y - 2 y d y

Simplify

= \int_{0}^{s i n (y)} - y d y

Take the constant out:

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

= - \int_{0}^{s i n (y)} y d y

Apply the Power Rule

: [\frac{y ^{2}}{2}]_{0}^{s i n (y)}

= - [\frac{y ^{2}}{2}]_{0}^{s i n (y)}

Compute the boundaries:

\frac{1}{2} sin^{2} (y)

= - \frac{1}{2} sin^{2} (y)

\int_{0}^{6} 16 x d x

\int_{0}^{5} ∣ 2 t - 5 ∣ d t

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

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

\int_{- 2}^{3} x - (x^{2} - 6) d x

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