What is the integral of 4sqrt(y-1) ?

The integral of 4sqrt(y-1) is 8/3 (y-1)^{3/2}+C

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integral of 4sqrt(y-1)

\int 4 y - 1 d y

\frac{8}{3} (y - 1)^{\frac{3}{2}} + C

Solution steps

\int 4 y - 1 d y

Take the constant out:

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

= 4 \cdot \int y - 1 d y

Apply u-substitution

= 4 \cdot \int u d u

Apply the Power Rule

= 4 \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = y - 1

= 4 \cdot \frac{2}{3} (y - 1)^{\frac{3}{2}}

Simplify

4 \cdot \frac{2}{3} (y - 1)^{\frac{3}{2}} : \frac{8}{3} (y - 1)^{\frac{3}{2}}

= \frac{8}{3} (y - 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{8}{3} (y - 1)^{\frac{3}{2}} + C

What is the integral of 4sqrt(y-1) ?
The integral of 4sqrt(y-1) is 8/3 (y-1)^{3/2}+C