What is the integral from x to 5 of (6(y-x))/(125) ?

The integral from x to 5 of (6(y-x))/(125) is (3(x^2-10x+25))/(125)

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integral from x to 5 of (6(y-x))/(125)

\int_{x}^{5} \frac{6 ( y - x )}{125} d y

\frac{3 ( x ^{2} - 10 x + 25 )}{125}

Solution steps

\int_{x}^{5} \frac{6 ( y - x )}{125} d y

Take the constant out:

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

= \frac{6}{125} \cdot \int_{x}^{5} - 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{6}{125} (- \int_{x}^{5} x d y + \int_{x}^{5} y d y)

\int_{x}^{5} x d y = x \cdot 5 - x^{2}

\int_{x}^{5} y d y = \frac{25 - x ^{2}}{2}

= \frac{6}{125} (- (x \cdot 5 - x^{2}) + \frac{25 - x ^{2}}{2})

Simplify

\frac{6}{125} (- (x \cdot 5 - x^{2}) + \frac{25 - x ^{2}}{2}) : \frac{3 ( x ^{2} - 10 x + 25 )}{125}

= \frac{3 ( x ^{2} - 10 x + 25 )}{125}

What is the integral from x to 5 of (6(y-x))/(125) ?
The integral from x to 5 of (6(y-x))/(125) is (3(x^2-10x+25))/(125)