What is the integral from a to b of g(y)-f(y) ?

The integral from a to b of g(y)-f(y) is (g(b^2-a^2))/2-(f(b^2-a^2))/2

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integral from a to b of g(y)-f(y)

\int_{a}^{b} g (y) - f (y) d y

\frac{g ( b ^{2} - a ^{2} )}{2} - \frac{f ( b ^{2} - a ^{2} )}{2}

Solution steps

\int_{a}^{b} g y - f 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_{a}^{b} g y d y - \int_{a}^{b} f y d y

\int_{a}^{b} g y d y = \frac{g ( b ^{2} - a ^{2} )}{2}

\int_{a}^{b} f y d y = \frac{f ( b ^{2} - a ^{2} )}{2}

= \frac{g ( b ^{2} - a ^{2} )}{2} - \frac{f ( b ^{2} - a ^{2} )}{2}

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What is the integral from a to b of g(y)-f(y) ?
The integral from a to b of g(y)-f(y) is (g(b^2-a^2))/2-(f(b^2-a^2))/2