Chapter 7.2, Problem 16E

To determine

  $\text{To verify that, }{\displaystyle \int f\left(u\right)du\ne }{\displaystyle \int f\left(u\right)dx}.$

Answer to Problem 16E

  $-\cos 4x+C\ne -\frac{1}{4}\cos 4x+C,\text{ the antiderivatives are not equal so the integrals are not equal}$ .

Explanation of Solution

Given information: $\text{The equation is},f\left(u\right)=\sin u\text{ and }u=4x\text{.}$

Calculation:

  $\begin{array}{l}\text{Find }{\displaystyle \int f\left(u\right)}du\text{ and then substitute in }u=4x,\\ {\displaystyle \int f\left(u\right)}du={\displaystyle \int \sin u}du\\ =-\cos u+C\\ =-\cos 4x+C\\ \text{Now, find }{\displaystyle \int f\left(u\right)}dx\text{ by substituting in }u=4x\text{ then integrating,}\\ {\displaystyle \int f\left(u\right)}dx={\displaystyle \int \sin u}dx\\ ={\displaystyle \int \sin 4x}dx\\ =-\frac{1}{4}\cos 4x+C\\ \text{therefore, }{\displaystyle \int f\left(u\right)}du=-\cos 4x+C\ne -\frac{1}{4}\cos 4x+C={\displaystyle \int f\left(u\right)}dx\\ \text{Hence, }-\cos 4x+C\ne -\frac{1}{4}\cos 4x+C,\text{ the antiderivatives are not equal so the integrals are not equal}\text{.}\end{array}$