What is the integral of 0.2x+8 ?

The integral of 0.2x+8 is 2.5(0.2x+8)^2+C

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integral of 0.2x+8

\int 0.2 x + 8 d x

2.5 (0.2 x + 8)^{2} + C

Solution steps

\int 0.2 x + 8 d x

Apply u-substitution

: \int \frac{u}{0.2} d u

= \int \frac{u}{0.2} d u

Take the constant out:

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

= \frac{1}{0.2} \cdot \int u d u

Simplify

= 5 \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

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

Substitute back

u = 0.2 x + 8

= 5 \cdot \frac{( 0.2 x + 8 ) ^{2}}{2}

Simplify

5 \cdot \frac{( 0.2 x + 8 ) ^{2}}{2} : 2.5 (0.2 x + 8)^{2}

= 2.5 (0.2 x + 8)^{2}

Add a constant to the solution

= 2.5 (0.2 x + 8)^{2} + C

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