Q. 17 - Calculus

Question

Show by exhibiting a counterexample that, in general, $\int f\left(x\right)g\left(x\right)dx\ne \left(\int f\left(x\right)dx\right)\left(\int g\left(x\right)dx\right)$. In other words, find two functions f and g so that the integral of their product is not equal to the product of their integrals. 

Step-by-Step Solution

Answer

The counterexample is $f\left(x\right)=x;g\left(x\right)=\frac{1}{x}.$

Step 1. Given information.

The given inequality is $\int f\left(x\right)g\left(x\right)dx\ne \left(\int f\left(x\right)dx\right)\left(\int g\left(x\right)dx\right).$

Step 2. Conclusion.

Let the two functions be,

$f\left(x\right)=x ; g\left(x\right)=\frac{1}{x}$

Now,

$$\begin{gathered} \int f\left(x\right)g\left(x\right)dx \\ =\int x\frac{1}{x}dx \\ =x+C \\ \left(\int f\left(x\right)dx\right)\left(\int g\left(x\right)dx\right)==\left(\int xdx\right)\left(\int \frac{1}{x}dx\right) \\ =\left(\frac{x^2}{2}+C\right)\left(\ln x+C^{\text{'}}\right) \\ Therefore, \\ \int f\left(x\right)g\left(x\right)dx\ne \left(\int f\left(x\right)dx\right)\left(\int g\left(x\right)dx\right) \end{gathered}$$