MATH241 Lecture Notes - Lecture 30: Antiderivative

Sketch the graph of f(x) = 2x3 on the interval [-2, 3]. Shade the signed area represented by f(x) dx, indicating the regions of positive and negative area. Without actually evaluating the definite integral, decide if the value of the integral is positive or negative. Briefly explain your answer. Using the formulas given in the text, and the basic properties of the integral, evaluate the definite integral (3x2 - x) dx. Suppose f (x)dx = 3 and f (x)dx = 2. Calculate f(x) dx f(x) dx f(x) dx 2f(x) dx Draw the graph of f(x) = 2x + 1. Evaluate f(x) dx by thinking of the integral as the sum the signed area of two triangular regions. Draw a graph of area represented by the definite integral (Hint: think about the graph of y = It's a familiar geometric shape.) Using geometry, determine the value of the definite integral. (Bonus question: worth 10 points. Total points for assignment not to exceed 100.) Prove that x2dx = b3/3 by computing the limit of the left-endpoint approximations.

Recall that if fis integrable on [a, b], then the Definite Integral of f from a to b is given by f(x)dx = lim f(xǐJAxk for any partition of [a, b] and any pointsx. To simplify these calculations, we use equally spaced grid points and right Riemann sums. That is, for each value of n, wlet AxkAand = a + kax, for k = 1,2, , n. Then as n → oo and Δ→ 0, k=1 In exercises #1-5, use the definition of the Definite Integral and right Riemann sums to evaluate the following definite integrals. Check your solutions using the Fundamental Theorem of Calculus. (1) x3)dx (2) (2x -4)dx (3) x?dx (4) (x3)dx (5) (2x2-1)dx The following Sums will be of help. Forn a positive integer and c a real number: c=cn k 1 k" = n(n+1)(2n + 1) nn 1)2

(1 point) View the following limit as a definite integral and then evaluate that integral by the Second Fundamental Theorem of Calculus. lim ntl 7n 7l The above limit is equal to

(1 point) Find the area of the surface generated by revolving the following curve about the axis: x = r cos t, y r sin t, 0 π. Area of the surface:

(1 point) The circle x-a cos , y = a sin t, 0 (doughnut). Find its surface area. Area of the torus: t 2π is revolved about the line x-b, 0く