**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:**

- 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.
- Repeat part (1) using left endpoints.
- Repeat part (1) using midpoints.
- 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.