What is the integral of 5sqrt(16-2x)-10 ?

The integral of 5sqrt(16-2x)-10 is -5/3 (16-2x)^{3/2}-10x+C

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integral of 5sqrt(16-2x)-10

\int 5 16 - 2 x - 10 d x

- \frac{5}{3} (16 - 2 x)^{\frac{3}{2}} - 10 x + C

Solution steps

\int 5 16 - 2 x - 10 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 5 16 - 2 x d x - \int 10 d x

\int 5 16 - 2 x d x = - \frac{5}{3} (16 - 2 x)^{\frac{3}{2}}

\int 10 d x = 10 x

= - \frac{5}{3} (16 - 2 x)^{\frac{3}{2}} - 10 x

Add a constant to the solution

= - \frac{5}{3} (16 - 2 x)^{\frac{3}{2}} - 10 x + C

f (x) = 7 e^{x}

\frac{d}{d x} (\frac{4 b}{a} x a^{2} - x^{2})

\frac{x ^{\frac{3}{2}}}{3}

(\frac{t ^{3}}{t ^{8} + 5})^{2}

y = x (x - 2)

What is the integral of 5sqrt(16-2x)-10 ?
The integral of 5sqrt(16-2x)-10 is -5/3 (16-2x)^{3/2}-10x+C