What is the integral of 30+0.05x ?

The integral of 30+0.05x is 10(0.05x+30)^2+C

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integral of 30+0.05x

\int 30 + 0.05 x d x

10 (0.05 x + 30)^{2} + C

Solution steps

\int 30 + 0.05 x d x

Apply u-substitution

: \int \frac{u}{0.05} d u

= \int \frac{u}{0.05} d u

Take the constant out:

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

= \frac{1}{0.05} \cdot \int u d u

Simplify

= 20 \cdot \int u d u

Apply the Power Rule

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

= 20 \cdot \frac{u ^{2}}{2}

Substitute back

u = 30 + 0.05 x

= 20 \cdot \frac{( 30 + 0.05 x ) ^{2}}{2}

Simplify

20 \cdot \frac{( 30 + 0.05 x ) ^{2}}{2} : 10 (0.05 x + 30)^{2}

= 10 (0.05 x + 30)^{2}

Add a constant to the solution

= 10 (0.05 x + 30)^{2} + C

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