Maths Using e and intergration

a curve has equation y = 6e^(3x) + x^(1/2) where x >= 0
The region bounded by the curve, the coordinate axis and the line with equation x = 1 is R
Calculate the area of R

I have no idea where to start with this one...
How to intergrate this? Do i need to find the maximum point to before I can intergrate?
Im really stuck!
Please someone walk me through this step by step!

To find the area of R, you integrate over the interval taking the upper bound minus the lower bound.

So, the interval is 0, 1.

Int_(0 to 1)_(6e^3x) + x^(1/2) dx

Split the integral at the + sign and take the constants out of the integrals

6Int_(0 to 1)_(e^3x)dx + Int(0 to 1)_(x^(1/2))dx

We have 6 times (1/3)e^(3x) evaluated from 0 to one plus (2/3)x^(3/2) evaluated from 0 to 1.
If we take the 1/3 out of the first term:
2(e^3x) evaluated from 0 to 1 + (2/3)x^(3/2) evaluated from 0 to 1.

2[ e^3(1) - e^3(0) ] + [2/3 - 0]
2[ e^3 - 1 ] + 2/3
2e^3 - 2 + 2/3
2e^3 - 4/3
2[ e^3 - 2/3 ]