What is the integral of (1/2 x-2)^4 ?

The integral of (1/2 x-2)^4 is (x^5}{80}-\frac{x^4)/4+2x^3-8x^2+16x+C

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integral of (1/2 x-2)^4

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

\frac{x ^{5}}{80} - \frac{x ^{4}}{4} + 2 x^{3} - 8 x^{2} + 16 x + C

Solution steps

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

Expand

(\frac{1}{2} x - 2)^{4} : \frac{x ^{4}}{16} - x^{3} + 6 x^{2} - 16 x + 16

= \int \frac{x ^{4}}{16} - x^{3} + 6 x^{2} - 16 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

= \int \frac{x ^{4}}{16} d x - \int x^{3} d x + \int 6 x^{2} d x - \int 16 x d x + \int 16 d x

\int \frac{x ^{4}}{16} d x = \frac{x ^{5}}{80}

\int x^{3} d x = \frac{x ^{4}}{4}

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

\int 16 x d x = 8 x^{2}

\int 16 d x = 16 x

= \frac{x ^{5}}{80} - \frac{x ^{4}}{4} + 2 x^{3} - 8 x^{2} + 16 x

Add a constant to the solution

= \frac{x ^{5}}{80} - \frac{x ^{4}}{4} + 2 x^{3} - 8 x^{2} + 16 x + C

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What is the integral of (1/2 x-2)^4 ?
The integral of (1/2 x-2)^4 is (x^5}{80}-\frac{x^4)/4+2x^3-8x^2+16x+C