What is the integral from 4 to x of ((x-4)^2)/(72) ?

The integral from 4 to x of ((x-4)^2)/(72) is 1/216 (x^3-12x^2+48x-64)

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integral from 4 to x of ((x-4)^2)/(72)

\int_{4}^{x} \frac{( x - 4 ) ^{2}}{72} d x

\frac{1}{216} (x^{3} - 12 x^{2} + 48 x - 64)

Solution steps

\int_{4}^{x} \frac{( x - 4 ) ^{2}}{72} d x

Take the constant out:

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

= \frac{1}{72} \cdot \int_{4}^{x} (x - 4)^{2} d x

Expand

(x - 4)^{2} : x^{2} - 8 x + 16

= \frac{1}{72} \cdot \int_{4}^{x} x^{2} - 8 x + 16 d x

Apply the Sum Rule:

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

= \frac{1}{72} (\int_{4}^{x} x^{2} d x - \int_{4}^{x} 8 x d x + \int_{4}^{x} 16 d x)

\int_{4}^{x} x^{2} d x = \frac{x ^{3} - 64}{3}

\int_{4}^{x} 8 x d x = 8 (\frac{x ^{2}}{2} - 8)

\int_{4}^{x} 16 d x = 16 x - 64

= \frac{1}{72} (\frac{x ^{3} - 64}{3} - 8 (\frac{x ^{2}}{2} - 8) + 16 x - 64)

Simplify

\frac{1}{72} (\frac{x ^{3} - 64}{3} - 8 (\frac{x ^{2}}{2} - 8) + 16 x - 64) : \frac{1}{216} (x^{3} - 12 x^{2} + 48 x - 64)

= \frac{1}{216} (x^{3} - 12 x^{2} + 48 x - 64)

x \to 1 lim (\frac{x}{x - 1} - \frac{1}{x})

\int \frac{x + 2}{x ^{2} + 4} d x

\frac{\partial}{\partial x} (7 x^{3 y})

\int \frac{x ^{7}}{( 1 + x ^{4} ) ^{2}} d x

(\frac{( e ^{6} )}{( e ^{4} )})^{- 1}

What is the integral from 4 to x of ((x-4)^2)/(72) ?
The integral from 4 to x of ((x-4)^2)/(72) is 1/216 (x^3-12x^2+48x-64)