What is the integral of 2*(x-6)^3 ?

The integral of 2*(x-6)^3 is 1/2 (x-6)^4+C

integral of 2*(x-6)^3

\int 2 \cdot (x - 6)^{3} d x

\frac{1}{2} (x - 6)^{4} + C

Solution steps

\int 2 (x - 6)^{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 - 6)^{3} d x

Apply u-substitution

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

Apply the Power Rule

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

Substitute back

u = x - 6

= 2 \cdot \frac{( x - 6 ) ^{4}}{4}

Simplify

2 \cdot \frac{( x - 6 ) ^{4}}{4} : \frac{1}{2} (x - 6)^{4}

= \frac{1}{2} (x - 6)^{4}

Add a constant to the solution

= \frac{1}{2} (x - 6)^{4} + C

t an g e n t f (x) = 7 + 4 x^{2} - 2 x^{3}, at x = a

d er i v a t i v e 4 e^{x^{- 3}}

x \to 0 lim (9 x^{2} - 4 x + 8)

\int sec^{2} (x) d x

y^{^{''}} + 1.5 y^{^{'}} + 0.037 y = 0, y (0) = 2, y^{^{'}} (0) = 1