if f(5)=14, f prime is continuous and integral(f '(x)) dx=18 from 5 to 7, what is the value of f(7)?
..please explain
..please explain
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By the FTC parts 1 and 2:
∫ f'(x) dx (from x=5 to 7) = f(7) - f(5).
However, since ∫ f'(x) dx (from x=5 to 7) = 18:
f(7) - f(5) = 18.
Therefore, solving for f(7) and plugging in the value of f(5) gives:
f(7) = 18 + f(5) = 18 + 14 = 32.
I hope this helps!
∫ f'(x) dx (from x=5 to 7) = f(7) - f(5).
However, since ∫ f'(x) dx (from x=5 to 7) = 18:
f(7) - f(5) = 18.
Therefore, solving for f(7) and plugging in the value of f(5) gives:
f(7) = 18 + f(5) = 18 + 14 = 32.
I hope this helps!
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integral of the derivative of a function IS the function itself
i.e. u just get rid of the prime sign (derivation sign ) and u have the result of the integration of the DERIVATIVE
better still, integral (f'(x)) dx = f(x) + c , for ANY indefinite f(x), with c a random constant
if the integral is definite, i.e. integral from a to b , then the result of the integration is f(a) - f(b)
hence, integral(f '(x)) dx = 18 from 5 to 7 => f(5) - f(7) = 18
u know f(5) = 14 => f(7) = f(5) - 18 = 14 -18 = -4
i.e. u just get rid of the prime sign (derivation sign ) and u have the result of the integration of the DERIVATIVE
better still, integral (f'(x)) dx = f(x) + c , for ANY indefinite f(x), with c a random constant
if the integral is definite, i.e. integral from a to b , then the result of the integration is f(a) - f(b)
hence, integral(f '(x)) dx = 18 from 5 to 7 => f(5) - f(7) = 18
u know f(5) = 14 => f(7) = f(5) - 18 = 14 -18 = -4