calculus help!?

Question

⌠3
[(3 x^2 + 4)/(x^2)] dx = ??
⌡2

Answer 1

We can break this up into two separate integrals, namely:

∫{2:3} 3x^2/x^2 dx + ∫{2:3} 4/x^2 dx

which we can simplify to:

∫{2:3} 3dx + ∫{2:3}4x^(-2) dx

I assume you can do these from here, but if not, we can integrate to get the following:

3x{2:3} + (-4/x){2:3}

= 3(3 - 2) -4(1/3 - 1/2)
= 3(1) -4(-1/6)
= 3 +2/3
= 11/3

--charlie

Answer 2

I'm guessing you're trying to say the integral from x = 2 to x = 3 of (3x^2 + 4)/(x^2) dx?

If so, let's first make the expression easier by saying

(3x^2 + 4)/(x^2) = 3x^2/x^2 + 4/x^2 = 3 + 4/x^2 = 3 + 4x^-2

So this becomes

int(3 + 4x^-2 dx) from x = 2 to x = 3
= [3x - 4x^-1] evaluated at x = 3 and x = 2

= [3(3) - 4/(3)] - [3(2) - 4/(2)]
= [9- (4/3)] - 4
= 5 - (4/3) = 11/3

Answer 3

The integral is 3x - 4/x
evaluated between 2 and 3 the answer is 11/3

Answer 4

11/3

[(3 x^2 + 4)/(x^2)] dx

=[3+4x^(-2) ]dx

integral value from 2 to 3 of
3x-4/x is 11/3

Answer 5

-14