What is the integral of ((0.5x-0.1)^4)/(0.4) ?

The integral of ((0.5x-0.1)^4)/(0.4) is (0.5x-0.1)^5+C

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integral of ((0.5x-0.1)^4)/(0.4)

\int \frac{( 0.5 x - 0.1 ) ^{4}}{0.4} d x

(0.5 x - 0.1)^{5} + C

Solution steps

\int \frac{( 0.5 x - 0.1 ) ^{4}}{0.4} d x

Take the constant out:

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

= \frac{1}{0.4} \cdot \int (0.5 x - 0.1)^{4} d x

Simplify

= 2.5 \cdot \int (0.5 x - 0.1)^{4} d x

Apply u-substitution

: \int \frac{u ^{4}}{0.5} d u

= 2.5 \cdot \int \frac{u ^{4}}{0.5} d u

Take the constant out:

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

= 2.5 \cdot \frac{1}{0.5} \cdot \int u^{4} d u

Simplify

= 2.5 \cdot 2 \cdot \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= 2.5 \cdot 2 \cdot \frac{u ^{5}}{5}

Substitute back

u = 0.5 x - 0.1

= 2.5 \cdot 2 \cdot \frac{( 0.5 x - 0.1 ) ^{5}}{5}

Simplify

2.5 \cdot 2 \cdot \frac{( 0.5 x - 0.1 ) ^{5}}{5} : (0.5 x - 0.1)^{5}

= (0.5 x - 0.1)^{5}

Add a constant to the solution

= (0.5 x - 0.1)^{5} + C

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What is the integral of ((0.5x-0.1)^4)/(0.4) ?
The integral of ((0.5x-0.1)^4)/(0.4) is (0.5x-0.1)^5+C