# If the solve the problem

Question:

If $f^{\prime}(x)=x+b, f(1)=5, f(2)=13$, find $f(x)$.

Solution:

Given $f^{\prime}(x)=x+b, f(1)=5$ and $f(2)=13$

On integrating the given equation, we have

$\int \mathrm{f}^{\prime}(\mathrm{x}) \mathrm{dx}=\int(\mathrm{x}+\mathrm{b}) \mathrm{dx}$

We know $\int \mathrm{f}^{\prime}(\mathrm{x}) \mathrm{dx}=\mathrm{f}(\mathrm{x})$

$\Rightarrow \mathrm{f}(\mathrm{x})=\int(\mathrm{x}+\mathrm{b}) \mathrm{d} \mathrm{x}$

$\Rightarrow \mathrm{f}(\mathrm{x})=\int \mathrm{x} \mathrm{dx}+\int \mathrm{bdx}$

$\Rightarrow \mathrm{f}(\mathrm{x})=\int \mathrm{xdx}+\mathrm{b} \int \mathrm{dx}$

Recall $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$ and $\int d x=x+c$

$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^{1+1}}{1+1}+\mathrm{b}(\mathrm{x})+\mathrm{c}$

$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^{2}}{2}+\mathrm{b} \mathrm{x}+\mathrm{c}$

On substituting $x=1$ in $f(x)$, we get

$\mathrm{f}(1)=\frac{1^{2}}{2}+\mathrm{b}(1)+\mathrm{c}$

$\Rightarrow 5=\frac{1}{2}+b+c$

$\Rightarrow 5-\frac{1}{2}=b+c$

$\Rightarrow b+c=\frac{9}{2} \ldots . .$  (1)

On substituting $x=2$ in $f(x)$, we get

$f(2)=\frac{2^{2}}{2}+b(2)+c$

$\Rightarrow 13=2+2 b+c$

$\Rightarrow 13-2=2 b+c$

$\Rightarrow 2 b+c=11 \ldots .(2)$

By subtracting equation (1) from equation (2), we have

$(2 b+c)-(b+c)=11-\frac{9}{2}$

$\Rightarrow 2 b+c-b-c=\frac{13}{2}$

$\therefore \mathrm{b}=\frac{13}{2}$

On substituting the value of $b$ in equation $(1)$, we get

$\frac{13}{2}+c=\frac{9}{2}$

$\Rightarrow \mathrm{c}=\frac{9}{2}-\frac{13}{2}$

$\therefore \mathrm{c}=-2$

On substituting the values of $b$ and $c$ in $f(x)$, we get

$f(x)=\frac{x^{2}}{2}+\frac{13}{2} x+(-2)$

$\therefore f(x)=\frac{x^{2}}{2}+\frac{13}{2} x-2$

Thus, $f(x)=\frac{x^{2}}{2}+\frac{13}{2} x-2$