Y 2 = nDeriv(Y 1 ,X,X)
Y 3 = nDeriv(Y 2 ,X,X)
Graph Y 3 and find its zero.
Make sure that only the equation sign in Y 3 is darkened.
Press 2nd and TRACE to access the CALC menu.
Select 2:zero and press ENTER, and a graph will appear.
Move the cursor anywhere to the left of where the second derivative graph crosses the x-axis,
and press ENTER to mark the left bound.
Move the cursor anywhere to the right of where the second derivative graph crosses the x-axis,
and press ENTER to mark the right bound.
Press ENTER when you are prompted for a guess.
The bottom of the graph will now show “Zero” and the value of x when y = 0.
x = 2.9999996 when y = 0
- C First, rewrite the integral as .
Now, we can use u-substitution to evaluate the integral. Let u = cos x. Then du = −sin x. We
can also change the limits of integration. The lower limit becomes cos 0 = 1 and the upper limit
becomes cos 1, which we leave alone. Now we perform the substitution, and we get
Evaluating the integral, we get −ln u = −ln(cos 1) + ln 1 = −ln(cos 1). This log is also
equal to ln(sec1).
- C In order to find the average value, we use the Mean Value Theorem for Integrals, which says
that the average value of f(x) on the interval [a, b] is f(x) dx.
Here we have
