If f'(x) = 10t / ∛(t – 2) and f(8) = –20, calculate f(x).

[s3a]

Homework Statement

Problem:
If f'(x) = 10t / ∛(t – 2) and f(8) = –20, calculate f(x).

Solution:
Let u = t – 2 ⇒ dx = du. Then f(x) = –20 + ∫_8^x [10t / ∛(t – 2)] dt = –20 + ∫_6^(x – 2) [10(u + 2) / ∛(u)] du = –20 + 10 ∫_6^(x – 2) [u^(2/3) + 2u^(–1/3)] du = 30 ∛[(x – 2)^2] + 6(x – 2)^(5/3) – 66 ∛(3) ∛(12) – 20

Additionally, the problem is attached as TheProblem.png, and the solution is attached as TheSolution.png.

Homework Equations

I'm not sure, but I think this has to do with the Fundamental Theorem of Calculus.

The Attempt at a Solution

I understand all the algebraic manipulations done; I'm just confused as to how the author went from the problem to the expression f(x) = –20 + ∫_8^x [10t / ∛(t – 2)] dt. Also, is it okay/valid that f'(x) (which is a function of x) = 10t / ∛(t – 2) (which is a function of t)?

Any help in clearing my confusions would be greatly appreciated!

 

[xiavatar]

Remember that when you integrate a function you have find the constant of integration. He just combined the two steps of finding the constant and integrating into one step, essentially.

 

[HallsofIvy]

The integral $\int_a^a h(t)dt= 0$ for any integrable function h. So that $\int_8^x h(t)dt$ gives a function that is 0 when x= 8. Knowing that f(8)= -20 means that $f(x)= -20+ \int_8^x h(t)dt$