# Recitation 18

Exercise: By applying the Mean Value Theorem to the function f(x)=x^{1/5} on the interval [32,33], show that 2 < \sqrt{33}{5} < 2.0125.

By the Mean Value Theorem, there is a c in the interval (32, 33) such that f'(c)=\sqrt{33}-\sqrt{32}. As f'(x)=\frac{1}{5}x^{-4/5}, f'(c)=\frac{1}{5c^{4/5}}. As 32 < c< 33, 0 < f'(c) < \frac{1}{5\times 32^{4/5}}=\frac{1}{80}. Thus 2 < \sqrt{33} < 2+\frac{1}{80}.

Exercise: Find the most general antiderivative of the function f(x)=e^x-(2/\sqrt{x}).

F(x)=e^x-4\sqrt{x}+C.

Exercise: Find f(x) such that f'(x)=2/(1+x^2), f(0)=-1.

f(x)=2\arctan(x)+C. As f(0)=C=-1, f(x)=2\arctan(x)-1.

Exercise:

1. Estimate the area under the graph of f(x)=1+x^2 from x=-1 to x=2 using three rectangles and right endpoints. Then improve your estimate by using six rectangles. Sketch the curve and the approximating rectangles.
2. Repeat part (1) using left endpoints.
3. Repeat part (1) using midpoints.
4. From your sketches in parts (1)-(3), which appears to be the best estimate?

Exercise: Find the Riemann sum for f(x)=x^3, -1\leq x\leq 1, if the partition points are -1, -0.5, 0, 0.5, 1 and the sample points are -1, -0.4, 0.2, 1.

The Riemann sum is (-1)^3\times 0.5+(-0.4)^3\times 0.5+(0.2)^3\times 0.5+1^3\times 0.5=-0.028.

Exercise: Use the Midpoint Rule with the given value of n to approximate the integral. Round the answer to four decimal places. \int_0^2\frac{x}{x+1}dx, n=5.

The partition points are 0, 0.4, 0.8, 1.2, 1.6, 2 and the sample points are 0.2, 0.6, 1, 1.4, 1.8. The Riemann sum is 0.4\times(\frac{0.2}{1.2}+\frac{0.6}{1.6}+\frac{1}{2}+\frac{1.4}{2.4}+\frac{1.8}{2.8})=0.7071.