What is the integral of (1-x/a)^3 ?

The integral of (1-x/a)^3 is x-3/a*(x^2)/2+(x^3)/(a^2)-(x^4)/(4a^3)+C

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integral of (1-x/a)^3

\int (1 - \frac{x}{a})^{3} d x

x - \frac{3}{a} \cdot \frac{x ^{2}}{2} + \frac{x ^{3}}{a ^{2}} - \frac{x ^{4}}{4 a ^{3}} + C

Solution steps

\int (1 - \frac{x}{a})^{3} d x

Expand

(1 - \frac{x}{a})^{3} : 1 - \frac{3 x}{a} + \frac{3 x ^{2}}{a ^{2}} - \frac{x ^{3}}{a ^{3}}

= \int 1 - \frac{3 x}{a} + \frac{3 x ^{2}}{a ^{2}} - \frac{x ^{3}}{a ^{3}} 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 1 d x - \int \frac{3 x}{a} d x + \int \frac{3 x ^{2}}{a ^{2}} d x - \int \frac{x ^{3}}{a ^{3}} d x

\int 1 d x = x

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

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

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

= x - \frac{3}{a} \cdot \frac{x ^{2}}{2} + \frac{x ^{3}}{a ^{2}} - \frac{x ^{4}}{4 a ^{3}}

Add a constant to the solution

= x - \frac{3}{a} \cdot \frac{x ^{2}}{2} + \frac{x ^{3}}{a ^{2}} - \frac{x ^{4}}{4 a ^{3}} + C

y^{^{''}} + 20 y^{^{'}} + y = 12

x \to 0 lim (3 + x^{2} sin (\frac{1}{x}))

y = \frac{1}{x - 4}, x = 8

\frac{( x ^{2} + 2 x - 6 )}{( x + 1 ) ^{2}}

((\frac{2}{s}) - (\frac{1}{s ^{3}}))^{2}

What is the integral of (1-x/a)^3 ?
The integral of (1-x/a)^3 is x-3/a*(x^2)/2+(x^3)/(a^2)-(x^4)/(4a^3)+C