What is the integral of y^4(1-y)^3 ?

The integral of y^4(1-y)^3 is (y^5)/5-(y^6)/2+(3y^7)/7-(y^8)/8+C

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integral of y^4(1-y)^3

integral

y^{4} (1 - y)^{3}

\frac{y ^{5}}{5} - \frac{y ^{6}}{2} + \frac{3 y ^{7}}{7} - \frac{y ^{8}}{8} + C

Solution steps

\int y^{4} (1 - y)^{3} d y

Expand

y^{4} (1 - y)^{3} : y^{4} - 3 y^{5} + 3 y^{6} - y^{7}

= \int y^{4} - 3 y^{5} + 3 y^{6} - y^{7} 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 3 y^{5} d y + \int 3 y^{6} d y - \int y^{7} d y

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

\int 3 y^{5} d y = \frac{y ^{6}}{2}

\int 3 y^{6} d y = \frac{3 y ^{7}}{7}

\int y^{7} d y = \frac{y ^{8}}{8}

= \frac{y ^{5}}{5} - \frac{y ^{6}}{2} + \frac{3 y ^{7}}{7} - \frac{y ^{8}}{8}

Add a constant to the solution

= \frac{y ^{5}}{5} - \frac{y ^{6}}{2} + \frac{3 y ^{7}}{7} - \frac{y ^{8}}{8} + C

\frac{1}{( x ) ^{2}}

e^{1} + 2 x

3 y + 7

\frac{1}{( x ^{2} + 2 x + 1 ) ^{2}}

\frac{4}{( 1 + x ^{2} )}

What is the integral of y^4(1-y)^3 ?
The integral of y^4(1-y)^3 is (y^5)/5-(y^6)/2+(3y^7)/7-(y^8)/8+C