# 4 if f(x)=7/sqrt(x^2 + 1) and f(4)=8.5, find f(2).

Explanation:

Step1: Usar la definición de integral

Sabemos que $f(x)-f(a)=\int_{a}^{x}f'(t)dt$. Aquí $a = 4$, $x = 2$, entonces $f(2)-f(4)=\int_{4}^{2}f'(t)dt=-\int_{2}^{4}f'(t)dt$.

Step2: Calcular la integral

Tenemos $f'(t)=\frac{7}{\sqrt{t^{2}+1}}$. Sabemos que $\int\frac{1}{\sqrt{x^{2}+1}}dx=\ln(x + \sqrt{x^{2}+1})+C$. Entonces $\int_{2}^{4}\frac{7}{\sqrt{t^{2}+1}}dt=7\int_{2}^{4}\frac{1}{\sqrt{t^{2}+1}}dt=7[\ln(t+\sqrt{t^{2}+1})]_{2}^{4}$.

Step3: Evaluar la integral definida

$7[\ln(t+\sqrt{t^{2}+1})]_{2}^{4}=7(\ln(4 + \sqrt{16 + 1})-\ln(2+\sqrt{4 + 1}))=7(\ln(4+\sqrt{17})-\ln(2+\sqrt{5}))$.

Step4: Encontrar $f(2)$

Como $f(2)-f(4)=-\int_{2}^{4}f'(t)dt$, entonces $f(2)=f(4)-7(\ln(4+\sqrt{17})-\ln(2+\sqrt{5}))$. Sustituyendo $f(4) = 8.5$, tenemos $f(2)=8.5-7(\ln(4+\sqrt{17})-\ln(2+\sqrt{5}))$. Calculando: [ \begin{align*} \ln(4+\sqrt{17})&\approx\ln(4 + 4.123)= \ln(8.123)\approx2.095\ \ln(2+\sqrt{5})&\approx\ln(2 + 2.236)=\ln(4.236)\approx1.444\ \ln(4+\sqrt{17})-\ln(2+\sqrt{5})&\approx2.095 - 1.444=0.651\ 7(\ln(4+\sqrt{17})-\ln(2+\sqrt{5}))&\approx7\times0.651 = 4.557\ f(2)&\approx8.5- 4.557=3.943 \end{align*} ]

Answer:

$f(2)\approx3.943$