I guess you mean
f(x) = x∫-6 [√(36-t^2) ] dt
in general ∫ (p^2-u^2)^.5 du = (1/2)[ u sqrt(p^2-u^2)+p^2(sin^-1(u/p)
here p = 6 , u = t
(1/2) [ t sqrt (36-t^2) + 36 sin^-1 (t/6) ] at t = x - at t = -6
(1/2) [ x sqrt(36-x^2) +36 sin^-1 (x/6)]
- (1/2) [-6sqrt(36-36) + 36 sin^-1(-1)]
= (1/2) [ x sqrt (36-x^2) +36 sin^-1(x/6) - 36 sin^-1 (-1)]
= (x/2) sqrt (36-x^2) + 18 sin^-1(x/6) - 18 sin^-1(-1)
well we can not take sqrt of a negative so x is between -6 and+ 6
when x = 0 I get
0 + 18(0 or pi) -18 (3 pi/4)
I am not sure I am going to get any of those answers, not sure I understood you question.
Can someone help me check my answers
1. Find the range of the function f(x) = x∫-6 √36-t^2 dt
- [-6, 0]
- [0, 6]
- [0, 9π] (my answer)
- [0, 18π]
2. Use the graph of f(t) = 2t + 2 on the interval [-1, 4] to write the function F(x), where f(x) = x∫-1 f(t) dt
a. F(x) = x^2 + 3x
b. F(x) = x^2 + 2x - 12 (my answer)
c. F(x) = x^2 + 2x - 3
d. F(x) = x^2 + 4x - 8
1 answer
1 answer