Find the area under the curve of f(x)=3x^2+2x in the interval 0.6

To find the area under the curve of the function f(x) = 3x^2 + 2x in the interval (0,6), we need to find the definite integral of the function over the interval (0,6).

We have:
\int_{0}^{6} (3x^2 + 2x) \, dx

Now, we integrate the function:
\int_{0}^{6} (3x^2 + 2x) \, dx = \left[ x^3 + x^2 \right]_{0}^{6}
= (6^3 + 6^2) - (0 + 0)
= (216 + 36) - 0
= 252

So, the area under the curve of f(x) = 3x^2 + 2x in the interval (0,6) is \boxed{252}.