What is the integral of 20+0.6sqrt(x) ?

The integral of 20+0.6sqrt(x) is 20x+0.4x^{3/2}+C

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integral of 20+0.6sqrt(x)

\int 20 + 0.6 x d x

20 x + 0.4 x^{\frac{3}{2}} + C

Solution steps

\int 20 + 0.6 x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 20 d x + \int 0.6 x d x

\int 20 d x = 20 x

\int 0.6 x d x = 0.4 x^{\frac{3}{2}}

= 20 x + 0.4 x^{\frac{3}{2}}

Add a constant to the solution

= 20 x + 0.4 x^{\frac{3}{2}} + C

(\frac{e}{16})^{x}

\int_{0}^{\frac{1}{2}} y^{2} (8 y) d y

y = (x^{2} - 1)

f (x) = 5 + 4 x^{2} - 2 x^{3}, at x = a

x \to \infty lim (- x + 2)

What is the integral of 20+0.6sqrt(x) ?
The integral of 20+0.6sqrt(x) is 20x+0.4x^{3/2}+C