Help with a calc problem?

Question

f(x)=sqrt(-40+14x-x^2) find the antiderivative satisfying the condition F(7)=3

Answer 1

sqrt(-40+14x-x²)
Substitution => x = 7:
\/(-40 + 14(7) - 7²) =
\/(-40+98-49) =
\/(-89 + 98) =
\/9 = 3 or -3
Answer: (x, y) = (7,3)
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Answer 2

Indefinite integral (antiderivative) of f(x) is

F(x) = 1/2 *[(x-7)*sqrt(-40+14 x - x^2) - 9 ArcSin((7-x)/3)] + C

For x = 7 we have F(x) = C, so C=3 if we want F(7) = 3.

Result:

F(x) = 1/2 *[(x-7)*sqrt(-40+14 x - x^2) - 9 ArcSin((7-x)/3)] + 3

Answer 3

rewrite f(x) to make integration a little easier

f(x)= 40^(1/2) + (14x)^1/2 -x^2^(1/2)
then integrate each part with respect to x
F(x)=x40^(1/2) + (2/3)(14^(1/2))x^(3/2)
+|x|(x/2)+c

then substitute 7 for x and find what c is(your integration constant)
F(7)=114.969+c=3
c=-111.969