What is the integral of (x/3-8)^4 ?

The integral of (x/3-8)^4 is (x^5}{405}-(8x^4)/(27)+\frac{128x^3)/9-(1024x^2)/3+4096x+C

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integral of (x/3-8)^4

\int (\frac{x}{3} - 8)^{4} d x

\frac{x ^{5}}{405} - \frac{8 x ^{4}}{27} + \frac{128 x ^{3}}{9} - \frac{1024 x ^{2}}{3} + 4096 x + C

Solution steps

\int (\frac{x}{3} - 8)^{4} d x

Expand

(\frac{x}{3} - 8)^{4} : \frac{x ^{4}}{81} - \frac{32 x ^{3}}{27} + \frac{128 x ^{2}}{3} - \frac{2048 x}{3} + 4096

= \int \frac{x ^{4}}{81} - \frac{32 x ^{3}}{27} + \frac{128 x ^{2}}{3} - \frac{2048 x}{3} + 4096 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}}{81} d x - \int \frac{32 x ^{3}}{27} d x + \int \frac{128 x ^{2}}{3} d x - \int \frac{2048 x}{3} d x + \int 4096 d x

\int \frac{x ^{4}}{81} d x = \frac{x ^{5}}{405}

\int \frac{32 x ^{3}}{27} d x = \frac{8 x ^{4}}{27}

\int \frac{128 x ^{2}}{3} d x = \frac{128 x ^{3}}{9}

\int \frac{2048 x}{3} d x = \frac{1024 x ^{2}}{3}

\int 4096 d x = 4096 x

= \frac{x ^{5}}{405} - \frac{8 x ^{4}}{27} + \frac{128 x ^{3}}{9} - \frac{1024 x ^{2}}{3} + 4096 x

Add a constant to the solution

= \frac{x ^{5}}{405} - \frac{8 x ^{4}}{27} + \frac{128 x ^{3}}{9} - \frac{1024 x ^{2}}{3} + 4096 x + C

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What is the integral of (x/3-8)^4 ?
The integral of (x/3-8)^4 is (x^5}{405}-(8x^4)/(27)+\frac{128x^3)/9-(1024x^2)/3+4096x+C