What is the integral of (3(x-3)^2)/(16) ?

The integral of (3(x-3)^2)/(16) is (x^3)/(16)-9/16 x^2+27/16 x+C

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integral of (3(x-3)^2)/(16)

\int \frac{3 ( x - 3 ) ^{2}}{16} d x

\frac{x ^{3}}{16} - \frac{9}{16} x^{2} + \frac{27}{16} x + C

Solution steps

\int \frac{3 ( x - 3 ) ^{2}}{16} d x

Take the constant out:

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

= \frac{3}{16} \cdot \int (x - 3)^{2} d x

Expand

(x - 3)^{2} : x^{2} - 6 x + 9

= \frac{3}{16} \cdot \int x^{2} - 6 x + 9 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{3}{16} (\int x^{2} d x - \int 6 x d x + \int 9 d x)

\int x^{2} d x = \frac{x ^{3}}{3}

\int 6 x d x = 3 x^{2}

\int 9 d x = 9 x

= \frac{3}{16} (\frac{x ^{3}}{3} - 3 x^{2} + 9 x)

Simplify

\frac{3}{16} (\frac{x ^{3}}{3} - 3 x^{2} + 9 x) : \frac{x ^{3}}{16} - \frac{9}{16} x^{2} + \frac{27}{16} x

= \frac{x ^{3}}{16} - \frac{9}{16} x^{2} + \frac{27}{16} x

Add a constant to the solution

= \frac{x ^{3}}{16} - \frac{9}{16} x^{2} + \frac{27}{16} x + C

\int (x + 7) d x

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

\frac{d}{d x} (3 x^{\frac{1}{3}} + 2 x^{2})

\int e^{3 x} sin (e^{x}) d x

\frac{d}{d x} (x + \frac{1}{3 x})

What is the integral of (3(x-3)^2)/(16) ?
The integral of (3(x-3)^2)/(16) is (x^3)/(16)-9/16 x^2+27/16 x+C