What is the integral from 0 to 1-y of 2 ?

The integral from 0 to 1-y of 2 is 2(-y+1)

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integral from 0 to 1-y of 2

\int_{0}^{1 - y} 2 d x

2 (- y + 1)

Solution steps

\int_{0}^{1 - y} 2 d x

Integral of a constant:

\int a d x = a x

= [2 x]_{0}^{1 - y}

Compute the boundaries:

2 (- y + 1)

= 2 (- y + 1)

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

\int_{0}^{\frac{π}{4}} 6 x + sec^{2} (x) - 2 d x

\int_{0}^{1} \frac{1}{1 - t ^{2}} d t

\int_{0}^{π} sin (t) cos (t) d t

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