Finding functions by integrating

In a nutshell

When given the gradient function $$f'(x)$$​, it is possible to find the function $$f(x)$$ by integrating. However, this gives $$f(x)$$ up to an unknown constant. This constant can be found if a specific point of $$f(x)$$​ is known.

Finding functions

To find a function $$y=f(x)$$​ given a gradient function $$\dfrac{dy}{dx}=f'(x)$$ and a point $$P(a,b)$$, follow this procedure.

procedure

Example 1

A function $$y=f(x)$$ has gradient function $$\dfrac{dy}{dx}=\dfrac{6}{\sqrt{x}}+12x$$. The function passes through the point $$P(1,5)$$. Find the function $$f(x)$$.

Integrate the gradient function to find $$f(x)$$ up to a constant:

$$\begin{aligned}f(x)&=\int \dfrac{dy}{dx}\, dx\\&=\int (\dfrac{6}{\sqrt{x}}+12x) \, dx \\&=\int 6x^{-\frac{1}{2}}\, dx + \int 12x^1\, dx +C\\&=\dfrac{6}{\frac{1}{2}}x^\frac12+\dfrac{12}{2}x^2 +C \\&=12x^{\frac12}+6x^2+C\end{aligned}$$​​

Therefore, $$y=12x^{\frac12}+6x^2+C$$.

Use the point $$P$$ to find the value of the constant $$C$$:

$$x=1,y=5\\ \begin{aligned}y&=12x^{\frac12}+6x^2+C\\5&=12(1)^{\frac12}+6(1)^2+C\\5&=12+6+C\\5&=18+C\\-13&=C\end{aligned}$$​​

Rewrite the function with the known value of $$C$$:

​$$\underline{f(x)=12x^{\frac12}+6x^2-13}$$​​