What is the integral of y^4-8y^3+16y^2 ?

The integral of y^4-8y^3+16y^2 is (y^5)/5-2y^4+(16y^3)/3+C

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integral of y^4-8y^3+16y^2

\int y^{4} - 8 y^{3} + 16 y^{2} d y

\frac{y ^{5}}{5} - 2 y^{4} + \frac{16 y ^{3}}{3} + C

Solution steps

\int y^{4} - 8 y^{3} + 16 y^{2} d y

Apply the Sum Rule:

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

= \int y^{4} d y - \int 8 y^{3} d y + \int 16 y^{2} d y

\int y^{4} d y = \frac{y ^{5}}{5}

\int 8 y^{3} d y = 2 y^{4}

\int 16 y^{2} d y = \frac{16 y ^{3}}{3}

= \frac{y ^{5}}{5} - 2 y^{4} + \frac{16 y ^{3}}{3}

Add a constant to the solution

= \frac{y ^{5}}{5} - 2 y^{4} + \frac{16 y ^{3}}{3} + C

\frac{\partial}{\partial y} (5 e^{y})

\frac{d}{d x} (tan^{2} (x^{2}))

\frac{d}{d x} (\frac{( x + 8 )}{x ^{2}})

n = 1 \sum \infty \frac{1}{4 + 2 ^{- n}}

What is the integral of y^4-8y^3+16y^2 ?
The integral of y^4-8y^3+16y^2 is (y^5)/5-2y^4+(16y^3)/3+C