What is the integral of 20+0.2x ?

The integral of 20+0.2x is 2.5(20+0.2x)^2+C

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integral of 20+0.2x

\int 20 + 0.2 x d x

2.5 (20 + 0.2 x)^{2} + C

Solution steps

\int 20 + 0.2 x d x

Apply u-substitution

= \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

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

Substitute back

u = 20 + 0.2 x

= 5 \cdot \frac{( 20 + 0.2 x ) ^{2}}{2}

Simplify

5 \cdot \frac{( 20 + 0.2 x ) ^{2}}{2} : 2.5 (20 + 0.2 x)^{2}

= 2.5 (20 + 0.2 x)^{2}

Add a constant to the solution

= 2.5 (20 + 0.2 x)^{2} + C