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

The integral of (1-x/4)^2 is -1/48 (4-x)^3+C

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integral of (1-x/4)^2

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

- \frac{1}{48} (4 - x)^{3} + C

Solution steps

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

Apply u-substitution

: \int - 4 u^{2} d u

= \int - 4 u^{2} d u

Take the constant out:

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

= - 4 \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= - 4 \cdot \frac{u ^{3}}{3}

Substitute back

u = 1 - \frac{x}{4}

= - 4 \cdot \frac{( 1 - \frac{x}{4} ) ^{3}}{3}

Simplify

- 4 \cdot \frac{( 1 - \frac{x}{4} ) ^{3}}{3} : - \frac{1}{48} (4 - x)^{3}

= - \frac{1}{48} (4 - x)^{3}

Add a constant to the solution

= - \frac{1}{48} (4 - x)^{3} + C

y^{^{'}} = - 2 x y^{3}

x \to \frac{3 π}{2} lim (sec (x))

x \to 0 lim (\frac{x ^{3}}{ln ∣ x ∣})

y^{^{'}} = y (x y^{3} + 5)

\int \frac{e ^{u}}{( 7 - e ^{u} ) ^{2}} d u

What is the integral of (1-x/4)^2 ?
The integral of (1-x/4)^2 is -1/48 (4-x)^3+C