Chapter 6.4, Problem 53E

To determine

To solve the problem using NINT function.

Answer to Problem 53E

The required value of x is $0.6989$ .

Explanation of Solution

Given information:

Find what value of $x$ does ${\displaystyle {\int }_0^x{e}^{-t^2}dt=0.6}$

Let $y={\displaystyle {\int }_0^x{e}^{-t^2}dt}$ , then the given integral becomes $y=0.6$ and it can be rewritten as $y-0.6=0$ .

So, to find the integral value, we enter the following syntax on a graphing calculator.

  $NINT\left(e^(-t^2),t,0,x\right)-0.6$

On graphing calculator, it shows that the graph touches the x -axis at $0.6989$ .

Conclusion:

Therefore, the value of x is $0.6989$ .