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

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

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integral of 2(x-3)^3

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

\frac{1}{2} (x - 3)^{4} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int u^{3} d u

= 2 \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= 2 \cdot \frac{u ^{4}}{4}

Substitute back

u = x - 3

= 2 \cdot \frac{( x - 3 ) ^{4}}{4}

Simplify

2 \cdot \frac{( x - 3 ) ^{4}}{4} : \frac{1}{2} (x - 3)^{4}

= \frac{1}{2} (x - 3)^{4}

Add a constant to the solution

= \frac{1}{2} (x - 3)^{4} + C

\frac{d ^{2} y}{d x ^{2}} + 4 y = sin (2)

\int x e^{y^{3}} d y

P = x^{3} - 5 x, (1, - 4)

cos (\frac{1 - e ^{8 x}}{1 + e ^{8 x}})

f (x) = (x^{2} + 3 x + 5) e^{6 x}