Question

Given: 5x^3+40=the integral from a to x of f(t)dt. The value of a is...
Answer: -2
Please explain how to come to this answer.

Answer 1

solve for x by taking cube root of each side

Answer 2

That works for this problem, but will that work for all these types of problems with integrals?

Answer 3

should.... also, it appears they want you to bring the negative OUT from under the root first. Because if it is under you could come up with imaginery #

Answer 4

you would only take the cube root of both sides if you are working with x^3
if working with x^2 then just take the square root

Answer 5

ok sounds good. Thank you. :)

Answer 6

\[\Large \int\limits_{a}^{x}f(t)dt = 5x^3 + 40\]Differentiate both sides:
f(t) = 15t^2
With f(t) = 15t^2, integrate f(t) between the limits: a and x. Equate it to 5x^3 + 40 and solve for a.

Answer 7

\[\Large \int\limits\limits_{a}^{x}15t^2dt = 5t^3 + k |_{a}^{x} = 5x^3 - 5a^3 = 5x^3 + 40\\5a^3 = -40; ~~a^3 = -8; ~~a = -2\]

Answer 8

Thank you I get it now :)

Answer 9

You are welcome.