Find f. f''(x) = x-2, x > 0, f(1) = 0, f(4) = 0.
Solution:
Given: f ''(x) = x-2
To find the original function, we need to integrate the given function twice.
Integrating on both the sides with respect to x.
⇒ ∫ f''(x)dx= ∫x-2dx
⇒ ∫ f''(x) dx = (x-2 + 1)/(-2 + 1) + C\(_1\)
⇒ f'(x) = x-1/(-1) + C\(_1\)
Integrating on both the sides with respect to x again
∫ f'(x) dx = -log(x) + C\(_1\)x + C\(_2\)
f(x) = -log(x) + C\(_1\)x + C\(_2\).
Given: f(1) =0
If x = 1, then f(1) = 0
0 = -log(1) + C\(_1\) + C\(_2\) --- (1)
Also f(4) = 0
0 = -log(4) + 4C\(_1\)+ C\(_2\) --- (2)
Solving (1) and (2) we get C\(_1\) and C\(_2\),
Subtracting (1) - (2)
0 = -log(1) + log(4) - 3C\(_1\)
3c1 = -log(1) + log(4) = log(4/1) {since log(a) - log(b) = log(a/b)}
C1 = log(4)/3
C1 = log(4)/3 and (2)
0 = -log(1) +(4/3)log(4) + C\(_2\)
log(1) - (4/3)log(4) = C\(_2\)
Replace C\(_1\) and C\(_2\) in equation in f(x)
f(x) = -log(x) + [log(4)]/3 + log(1) - (4/3)log(4)
f(x) = -log(x) + log(1) - log(4).
Find f. f''(x) = x - 2, x > 0, f(1) = 0, f(4) = 0.
Summary:
If f ''(x) = x-2, x > 0, f(1) = 0, f(4) = 0 then r ⇒ f(x) = -log(x) + log(1) - log(4).
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