What is the integral from x to 1 of 6/7 x+12/7 y ?

The integral from x to 1 of 6/7 x+12/7 y is (6(-x^2+1))/7+(6x-6x^2)/7

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integral from x to 1 of 6/7 x+12/7 y

\int_{x}^{1} \frac{6}{7} x + \frac{12}{7} y d y

\frac{6 ( - x ^{2} + 1 )}{7} + \frac{6 x - 6 x ^{2}}{7}

Solution steps

\int_{x}^{1} \frac{6}{7} x + \frac{12}{7} 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} \frac{12}{7} y d y + \int_{x}^{1} \frac{6}{7} x d y

\int_{x}^{1} \frac{12}{7} y d y = \frac{6 ( - x ^{2} + 1 )}{7}

\int_{x}^{1} \frac{6}{7} x d y = \frac{6 x - 6 x ^{2}}{7}

= \frac{6 ( - x ^{2} + 1 )}{7} + \frac{6 x - 6 x ^{2}}{7}

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What is the integral from x to 1 of 6/7 x+12/7 y ?
The integral from x to 1 of 6/7 x+12/7 y is (6(-x^2+1))/7+(6x-6x^2)/7