Question

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Suppose f(x) = x² + 4x and g(x) is an anti derivative of f(x).

If g(5) = 7, then find the value of g(1).​

Answer 1

Answer:

the answer is 5891.34...50

Answer 2

Answer:

Answer

xg(f(x))f

′

(g(x))g

′

(x)=f(g(x))g

′

(f(x))f

′

(x)

x \dfrac {f'(g(x) g'(x) }{f(g(x) }=\dfrac{ g'(f(x) f(x)}{g(f(x)) }

integrating the equation , as we know integration of $$ \dfrac {f'(g(x) g'(x) }{f(g(x) } =lnf(g(x)) andintegrationof \dfrac {g'(f(x) g'(x) }{g(f(x) }= ln g(f(x)$$

after integration we get

xln f(g(x)) - integration of (ln f(g(x)) dx =lng(f(x)1 equation

integration from zero to a of f(g(x))dx = \dfrac {(1-e^-2a }{ 2}

now differentiating this equation

f(g(x))=e^-2aequation 2

now putting equation 2 in equation 1 we get

xln(e^-2x)- integration ln(e^-2x) dx = ln(g(f( x))

-2x \times x -x^2 = ln g(f(X))

lng(f(X))= -x^2

g(f(x))= e^(-x^2)

g(f(4)) = e^(-4 *4)

so k= 4

Explanation:

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