Step 1

We have given \(\displaystyle{F}\text{}{x}{)}={1},{F}'{\left({0}\right)}={3},{F}{\left({0}\right)}={4}\)

We have to find the function F.

Step 2

We have

\(\displaystyle{F}\text{}{x}{)}={1}\)

Take integration on both sides

\(\displaystyle\int{F}\text{}{x}{)}{\left.{d}{x}\right.}=\int{\left.{d}{x}\right.}\)

\(\displaystyle\Rightarrow{F}'{\left({x}\right)}={x}+{C}\)

We have given

\(\displaystyle{F}'{\left({0}\right)}={3}\)

Then, \(\displaystyle{F}'{\left({0}\right)}={3}={0}+{C}\)

\(\displaystyle\Rightarrow{C}={3}\)

So, \(\displaystyle\Rightarrow{F}'{\left({x}\right)}={x}+{3}\) (i)

Taking integration on both sides of equation (i) we get,

\(\displaystyle\int{F}'{\left({x}\right)}{\left.{d}{x}\right.}=\int{x}{\left.{d}{x}\right.}+\int{3}{\left.{d}{x}\right.}\)

\(\displaystyle\Rightarrow{F}{\left({x}\right)}={\frac{{{x}^{{{2}}}}}{{{2}}}}+{3}{x}+{D}\)

And also we have \(\displaystyle{F}{\left({0}\right)}={4}\)

Then, \(\displaystyle{F}{\left({0}\right)}={4}={\frac{{{0}^{{{2}}}}}{{{2}}}}+{0}+{D}\)

\(\displaystyle\Rightarrow{D}={4}\)

Therefore, function is

\(\displaystyle{F}{\left({x}\right)}={\frac{{{x}^{{{2}}}}}{{{2}}}}+{3}{x}+{4}\)

We have given \(\displaystyle{F}\text{}{x}{)}={1},{F}'{\left({0}\right)}={3},{F}{\left({0}\right)}={4}\)

We have to find the function F.

Step 2

We have

\(\displaystyle{F}\text{}{x}{)}={1}\)

Take integration on both sides

\(\displaystyle\int{F}\text{}{x}{)}{\left.{d}{x}\right.}=\int{\left.{d}{x}\right.}\)

\(\displaystyle\Rightarrow{F}'{\left({x}\right)}={x}+{C}\)

We have given

\(\displaystyle{F}'{\left({0}\right)}={3}\)

Then, \(\displaystyle{F}'{\left({0}\right)}={3}={0}+{C}\)

\(\displaystyle\Rightarrow{C}={3}\)

So, \(\displaystyle\Rightarrow{F}'{\left({x}\right)}={x}+{3}\) (i)

Taking integration on both sides of equation (i) we get,

\(\displaystyle\int{F}'{\left({x}\right)}{\left.{d}{x}\right.}=\int{x}{\left.{d}{x}\right.}+\int{3}{\left.{d}{x}\right.}\)

\(\displaystyle\Rightarrow{F}{\left({x}\right)}={\frac{{{x}^{{{2}}}}}{{{2}}}}+{3}{x}+{D}\)

And also we have \(\displaystyle{F}{\left({0}\right)}={4}\)

Then, \(\displaystyle{F}{\left({0}\right)}={4}={\frac{{{0}^{{{2}}}}}{{{2}}}}+{0}+{D}\)

\(\displaystyle\Rightarrow{D}={4}\)

Therefore, function is

\(\displaystyle{F}{\left({x}\right)}={\frac{{{x}^{{{2}}}}}{{{2}}}}+{3}{x}+{4}\)