What is the integral from x to 1 of (x+y) ?

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

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integral from x to 1 of (x+y)

\int_{x}^{1} (x + y) d y

x - x^{2} + \frac{1 - x ^{2}}{2}

Solution steps

\int_{x}^{1} x + y d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{x}^{1} x d y + \int_{x}^{1} y d y

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

\int_{x}^{1} y d y = \frac{1 - x ^{2}}{2}

= x - x^{2} + \frac{1 - x ^{2}}{2}

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