Liwen Guo

2022-06-30

RizerMix

To find the function f given the derivative ${f}^{\prime }\left(t\right)=\frac{x+1}{\sqrt{x}}$ and the initial condition f(1) = 5, we can integrate the derivative with respect to x to obtain f(x).
Integrate f'(t) with respect to x:
$\int \frac{x+1}{\sqrt{x}}\phantom{\rule{0.167em}{0ex}}dx$
Simplify the integrand:
$\int \left(x+1\right){x}^{-1/2}\phantom{\rule{0.167em}{0ex}}dx$
Distribute the integral and split it into two separate integrals:
$\int x·{x}^{-1/2}\phantom{\rule{0.167em}{0ex}}dx+\int 1·{x}^{-1/2}\phantom{\rule{0.167em}{0ex}}dx$
Simplify each integral separately:
$\int x·{x}^{-1/2}\phantom{\rule{0.167em}{0ex}}dx=\int {x}^{1/2}\phantom{\rule{0.167em}{0ex}}dx=\frac{2}{3}{x}^{3/2}+{C}_{1}$
$\int 1·{x}^{-1/2}\phantom{\rule{0.167em}{0ex}}dx=\int {x}^{-1/2}\phantom{\rule{0.167em}{0ex}}dx=2{x}^{1/2}+{C}_{2}$
Combine the results and include the constants of integration:
$f\left(x\right)=\frac{2}{3}{x}^{3/2}+{C}_{1}+2{x}^{1/2}+{C}_{2}$
Apply the initial condition f(1) = 5 to find the values of the constants C1 and C2:
$f\left(1\right)=\frac{2}{3}\left({1}^{3/2}\right)+{C}_{1}+2\left({1}^{1/2}\right)+{C}_{2}=\frac{2}{3}+{C}_{1}+2+{C}_{2}=5$
Solve for the constants C1 and C2:
$\frac{2}{3}+{C}_{1}+2+{C}_{2}=5$
Combine like terms:
${C}_{1}+{C}_{2}+\frac{8}{3}=5$
Subtract $\frac{8}{3}$ from both sides:
${C}_{1}+{C}_{2}=5-\frac{8}{3}$
${C}_{1}+{C}_{2}=\frac{7}{3}$
Substitute the values of C1 and C2 back into the equation for f(x):
$f\left(x\right)=\frac{2}{3}{x}^{3/2}+\frac{7}{3}$
Therefore, the function f(x) is $f\left(x\right)=\frac{2}{3}{x}^{3/2}+\frac{7}{3}$.

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