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MAS110 Problems for Chapter 7: Integration 1. By slicing the area into N thin pieces approximated by rectangles, and proceeding as for RX in the first integration lecture, find an alternative proof that 0 x dx = X 2 /2. RX 0 x2 dx RX 3 2. Strictly speaking, in the lecture we found an overestimate: 0 x2 dx < X6 1 + N1 2 + N1 . Some hand-waving was required to believe that as N → ∞ the limit of the overestimate is the true area. Here is a more careful approach. rectangles that cut underneath the curve By Rusing X 2 1 1 X3 obtain an underestimate: 6 1 − N 2 − N < 0 x dx. The true area is definitely sandwiched between the underestimate and the overestimate. Now what happens as N → ∞? 3. Here is another example of using Fermat’s method of subdividing an interval to calculate integrals. Let X > 0 be a real number. (a) Sketch a picture to illustrate an overestimate the area of the region between the x-axis and the curve y = x1/2 over the interval [0, X] by using the points . . . , a3 X, a2 X, aX, X, for some 0 < a < 1. Calculate the overestimate. RX (b) Letting a → 1 calculate 0 x1/2 dx. RX (c) If you want to you can work out 0 xm/n dx for integers m, n 6= 0 with m/n > −1 using the same method. 4. Let n be a positive integer and X > 0 a real number. Calculate your answer geometrically in terms of two areas. RX 0 xn dx + R Xn 0 x1/n dx. Explain 5. Assume all functions that appear Zare sufficiently well-behaved. What does the Fundamental The d x orem of Calculus say about f (t) dt ? Combining with the chain rule, show that dx 0 d dx d Obtain a similar relation for dx Z Z h(x) f (t) dt = f (h(x))h0 (x). 0 h(x) f (t) dt . g(x) Z Z 6. Give an example to illustrate the fact that in general Z 7. Calculate 2π Z x sin x dx and 0 u(x)v(x) dx 6= Z u(x) dx v(x) dx . 2π x2 cos 2x dx. 0 8. We knowZhow to integrate cos2 x using the formula cos2 x = (1 + cos 2x)/2. As an alternative, π evaluate cos2 x dx by writing cos2 x = cos x cos x, and integrating by parts. 0 1 9. Let f and g be functions selected from the set {1, cos x, sin x, cos 2x, sin 2x, cos 3x, sin 3x, . . .}. Integrating by parts, and using the formulas for sin2 θ Z 2π 0 f (x)g(x) dx = 2π 0 π 1 Z x2 10. Differentiate e . Evaluate the integrals and cos2 θ, show that if f 6= g; if f = g = 1; if f = g 6= 1. Z x2 0 Z 1 11. Evaluate the integrals 0 x2 dx, 1 + x3 Z 12. Integrating by parts, evaluate Z Z 0 1 1 2 x3 ex dx. xe dx and 0 x2 dx and 1 + x2 Z 1 0 x3 dx. 1 + x2 tan−1 x dx. π ex cos x dx. 13. Evaluate 0 d (sec x) = sec x tan x. By 14. Show that if −π/2 < x < π/2 then sec x + tan x > 0. ZShow that dx making the substitution u = sec x + tan x, show that sec x dx = ln(sec x + tan x) + C (if −π/2 < x < π/2). d dx 2 3 2 (tan x) = sec Z x? By writing sec x = sec x sec x and integrating by parts, evaluate the indefinite integral sec3 x dx. 15. Why is Z 16. By making a trigonometric substitution, show that 0 Z 2 17. Evaluate x2 1 Z x √ √ 1 dt = ln(x + x2 + 1). 1 + t2 1 dx. + 2x 1 1 √ dx. At some point you might like to use a trigonometric substitution x2 + 2x 0 involving sec θ. 18. Evaluate 19. Evaluate the following improper integrals. Z ∞ Z ∞ ln(x) 3 −x4 dx; (b) (a) xe dx; x2 0 1 2 Z (c) 0 ∞ xn e−x dx where n ∈ N.