## Calculus: Early Transcendentals 8th Edition

Given:$\int_{a}^{b}[f(x)g(x)]dx=(\int_{a}^{b}f(x))dx)(\int_{a}^{b}g(x)dx$ Consider $f(x)=x,g(x)=x^{2}, a=0, b=1$ Take Left Side: $\int_{0}^{1}(x)(x^{2} )dx=\int_{0}^{1}(x^{3} )dx$ $=[\frac{x^{4}}{4}]_{0}^{1}$ $=\frac{1}{4}$ Take right side: $(\int_{0}^{1}(x))dx)(\int_{0}^{1}(x^{2})dx=[\frac{x^{2}}{2}]_{0}^{1}[\frac{x^{3}}{3}]_{0}^{1}$ $=\frac{1}{2} \times \frac{1}{3}$ $=\frac{1}{6}$ Thus, $\frac{1}{4}\ne\frac{1}{6}$ Hence, $\int_{a}^{b}[f(x)g(x)]dx\ne(\int_{a}^{b}f(x))dx)(\int_{a}^{b}g(x)dx$ The given statement is false.