Chapter 6.3, Problem 21E

To determine

ToEvaluate:The integral ${\displaystyle \underset{0}{\overset{1}{\int }}{e}^{x}dx}$ using the antiderivative by interpreting the integrand as the rate of change of a quantity.

Answer to Problem 21E

  ${\displaystyle \underset{0}{\overset{1}{\int }}{e}^{x}dx}=e-1$

Explanation of Solution

Given:

The integral ${\displaystyle \underset{0}{\overset{1}{\int }}{e}^{x}dx}$

Concept Used:

To evaluate the definite integral of $f$ from $a$ to any number $x$ . Simply compute $F\left(x\right)-F\left(a\right)$ , where $F$ is any antiderivative of $f$ .

That is

  ${\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)dt}=F\left(x\right)-F\left(a\right)$

Calculation:

Given the integral ${\displaystyle \underset{0}{\overset{1}{\int }}{e}^{x}dx}$

Since $e^x$ is the rate of change of the quantity $F\left(x\right)=e$

That is

  $F\text{'}\left(x\right)=e$

Evaluating theintegral ${\displaystyle \underset{0}{\overset{1}{\int }}{e}^{x}dx}$ , using the formula, ${\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)dt}=F\left(x\right)-F\left(a\right)$

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{1}{\int }}{e}^{x}dx}=e^1-e^0\\ =e-1\end{array}$

Therefore, ${\displaystyle \underset{0}{\overset{1}{\int }}{e}^{x}dx}=e-1$

Conclusion:

  ${\displaystyle \underset{0}{\overset{1}{\int }}{e}^{x}dx}=e-1$