How do you find the definite integral of #int (4x^2+2)dx# from #[0,3]#?

1 Answer

Hi!

To solve most definite integral problems, the process is pretty well the same:

1) Integrate the function - that is, get it's "anti-derivative"
2) Evaluate at the upper limit
3) Evaluate at the lower limit
4) Evaluate difference between the upper and lower limit

Explanation:

In your example, you must integrate your function, then evaluate it at 0, then at 3, then evaluate the difference.

I'll run through it below:

#int_0^3(4x^2+2)dx# on # [0, 3]#

#4/3x^3 + 2x -> # Note: you omit the constant of integration "+ c" as it cannot be evaluated without initial conditions.

Also, you'd normally put a vertical bar after the integrated function with the lower limit at the bottom, and the upper limit at the top. This simply means, "Evaluate the function from the lower limit "a" to the upper limit "b". I'd use that notation however, I'm not sure how to put it in, haha! :)

Anyway, now you can evaluate your integrated function at both limits:

# = (4/3(3)^3 + 2(3)) - (4/3(0)^3 + 2(0)) #
# = 42 #

That's it! That's how you evaluate definite integrals! Hopefully everything was clear and helpful! :)