What is the integral from x to 2 of (6(y-x))/8 ?

The integral from x to 2 of (6(y-x))/8 is (3(x^2-4x+4))/8

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integral from x to 2 of (6(y-x))/8

\int_{x}^{2} \frac{6 ( y - x )}{8} d y

\frac{3 ( x ^{2} - 4 x + 4 )}{8}

Solution steps

\int_{x}^{2} \frac{6 ( y - x )}{8} d y

Take the constant out:

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

= \frac{6}{8} \cdot \int_{x}^{2} - x + y d y

Cancel the common factor:

2

= \frac{3}{4} \cdot \int_{x}^{2} - 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

= \frac{3}{4} (- \int_{x}^{2} x d y + \int_{x}^{2} y d y)

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

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

= \frac{3}{4} (- (x \cdot 2 - x^{2}) + 2 - \frac{x ^{2}}{2})

Simplify

\frac{3}{4} (- (x \cdot 2 - x^{2}) + 2 - \frac{x ^{2}}{2}) : \frac{3 ( x ^{2} - 4 x + 4 )}{8}

= \frac{3 ( x ^{2} - 4 x + 4 )}{8}

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What is the integral from x to 2 of (6(y-x))/8 ?
The integral from x to 2 of (6(y-x))/8 is (3(x^2-4x+4))/8