Question

Determine whether the statement is true or false if It is true, explain why, if it is false, explain why or give an example that disproves the statement

If f and g are continuous on(a,b)then

\(\int\limits_a^b {\left( {f(x)g(x)} \right)dx = } \left( {\int\limits_a^b {f(x)dx} } \right)\left( {\int\limits_a^b {g(x)dx} } \right)\)

Short answer

Hence proved the given statement is false\(\int\limits_a^b {\left( {f(x)g(x)} \right)dx \ne } \left( {\int\limits_a^b {f(x)dx} } \right)\left( {\int\limits_a^b {g(x)dx} } \right)\).

Step-by-step solution

Step 1: Solution of LHS of the equation

Let consider some examples to expression and apply the limit a is 0 and b is 1

\(\begin{array}{l}f(x) = x\\g(x) = {x^2}\\\int_0^1 {(x)} \left( {{x^2}} \right)dx = \int_0^1 {{x^3}} dx\end{array}\)

\(\left( {\frac{{{x^4}}}{4}} \right)_0^1 = \frac{1}{4}\)

Step 2: Solution of RHS of the equation

\(\begin{array}{l}\left( {\int_0^1 x dx} \right)\left( {\int_0^1 {{x^2}} dx} \right) = \left( {\frac{{{x^2}}}{2}} \right)_0^1\left( {\frac{{{x^3}}}{3}} \right)_0^1\\ = \left( {\frac{1}{2}} \right)\left( {\frac{1}{3}} \right)\\ = \frac{1}{6}\end{array}\)

Step 3: Final Proof

Consider from LHS and RHS solutions both sides are not equal

Hence the equation is proved and given statement is false.

\(\begin{array}{l}\int\limits_a^b {\left( {f(x)g(x)} \right)dx \ne } \left( {\int\limits_a^b {f(x)dx} } \right)\left( {\int\limits_a^b {g(x)dx} } \right)\\\frac{1}{4} \ne \frac{1}{6}\end{array}\)