What is the integral of 10(y^4-y+1) ?

The integral of 10(y^4-y+1) is 10((y^5)/5-(y^2)/2+y)+C

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integral of 10(y^4-y+1)

\int 10 (y^{4} - y + 1) d y

10 (\frac{y ^{5}}{5} - \frac{y ^{2}}{2} + y) + C

Solution steps

\int 10 (y^{4} - y + 1) d y

Take the constant out:

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

= 10 \cdot \int y^{4} - y + 1 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

= 10 (\int y^{4} d y - \int y d y + \int 1 d y)

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

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

\int 1 d y = y

= 10 (\frac{y ^{5}}{5} - \frac{y ^{2}}{2} + y)

Add a constant to the solution

= 10 (\frac{y ^{5}}{5} - \frac{y ^{2}}{2} + y) + C

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What is the integral of 10(y^4-y+1) ?
The integral of 10(y^4-y+1) is 10((y^5)/5-(y^2)/2+y)+C