if the integral ( [ )
([ f(t) ,t, 0, 8) = k
Find [ (3x^2f(x^3), x, 0, 2)
oh the answer is k they said the integral of the function i said to find from 0 to 2 is equal to the integral from 0 to 8
WHYYY
Why would you say that? I understand you use u subsitution u= x^3 and then eventually you get an integral from 0 2 of the function but i dnt understand why the integral from 0 to 2 and the integral from 0 to 8 would be the same?
To find the integral of (3x^2)f(x^3) with respect to x from 0 to 2, we can apply a change of variable to simplify the integral. Let's substitute u = x^3, so the differential du = 3x^2 dx. This gives us:
∫(3x^2)f(x^3) dx = ∫f(u) du
Now, we need to determine the bounds of integration in terms of u. When x = 0, u = (0)^3 = 0, and when x = 2, u = (2)^3 = 8. So, the new integral becomes:
∫f(u) du, where u ranges from 0 to 8.
Given that the integral of f(t) with respect to t from 0 to 8 is equal to k, we can rewrite the above integral as:
∫f(u) du = k
Therefore, the value of the integral is k.