Indefinite Integral: int square root of y (y^2 -y+1)dy

[cmnalo]

Find the indefinite integral

∫square root of y (y^2 -y+1)dy

∫y^1/2 (∫y^2dy -∫ydy + 1∫dy

2/3y^3/2 + c (1/3y^3 - 1/2y^2 + y +c)

I don't know if I'm doing this right or what the next step would be. Any help would be appreciated.

Answer:
2/7)y^7/2-(2/5)y^5/2+(2/3)y^3/2+C

 

[pka]

It is very hard to follow you notation. You may want to learn to use TeX.
However, the integral is additive but not multiplicative!
\(\displaystyle \L\int {f(x)g(x)dx \not= } \left( {\int {f(x)dx} } \right)\left( {\int {g(x)dx} } \right)\)

 

[stapel]

Did your book (or instructor) really tell you that you could split the integral of a product into a product of integrals? :shock:

Try the following:

. . . . .\(\displaystyle \L \int\,\left(y\, -\, y^{\frac{3}{2}}\, +\, y^{\frac{1}{2}}\right)\, dy\)

If you get stuck, please reply showing all of your steps, beginning with the above restatement of the integral. Thank you.

Eliz.

 

[cmnalo]

Eliz-

So, did you distribute the y^1/2 into the parenthesis?

i.e.

∫[(y^1/2 * y^2)-(y^1/2 * y) + (y^1/2 * 1)] ?

 

[stapel]

...did you distribute the y^1/2 into the parenthesis?

The integrand was simplified by using the exponent rules you learned back in algebra.

Eliz.