HOMEWORK SOLUTIONS MATH 1910 Sections 5.1, 5.2 Fall 2016

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1 HOMEWORK SOLUTIOS MATH 9 Sections.,. Fall 6 Problem.. An ostrich runs with velocit km/h for minutes min), km/h for min, and km/h for another minute. Compute the total distance traveled and indicate with a graph how this quantit can be interpreted as an area. SOLUTIO. The total distance traveled b the ostrich is ) ) + 6 ) ) + 6 ) ) = = 9 km This distance is the area under the graph below which shows the ostrich s velocit as a function of time. Problem.. Compute R and L over [, ] using the following values: f) SOLUTIO. We have = =.. Hence we have L = ) =.) = 6 and R = ) =.) =...

2 Problem.. Use linearit and formulas )-) to rewrite and evaluate the sum m= + m ) SOLUTIO. m= + m ) = m= = + m + 9 m ) m= + = + m= m + 9 m m= + ) ) 6 = 7... Problem..86 Prove that for an function f on [a, b], R L = b a fb) fa)). SOLUTIO. Let = b a)/ and partition the interval [a, b] into equal-length subintervals with endpoints a = < < < < = b. We have L = f k ) and R = k= R L = = k= f k ), so k= ) f k ) f k ) f ) + k= f k ) f ) + k= = f ) f )) = b a k= f k ) )) fb) fa))...86

3 Problem..88 Use [the previous problem] to show that if f is positive and monotonic, then the area A under its graph over [a, b] satisfies R A b a fb) fa). SOLUTIO. Suppose f is positive and increasing. Then R is an over-approimation of the area under the graph of f, and L is an under-approimation. That is, we have L A R. Subtracting R from all three parts of that inequalit ields and negating all three parts gives From the previous problem we have L R A R, R A R L. R A b a and taking absolute values gives the desired result. fb) fa)), R A b a fb) fa)) Since R A and fb) fa) are positive, the are equal to their own absolute values.) The case in which f is decreasing is similar...88

4 SECTIO. The Definite Integral Problem..7.. Draw a graph of the signed area represented b the integral and compute it using geometr: 7. d d. solution The region bounded b the graph of = and the -ais over the interval [, ] is one-quarter of a circle of radius. Hence, SOLUTIO. The region below = over [, ] is a quarter of the circle of radius centered at the origin: d = π) = π SECTIO.. The Definite Integral The region bounded b the graph of = 8andthe-ais over the interval [, ] consists of a triangle below the ais with base and height and a triangle above the ais with base and height 8. Hence, 8)d = )) + 8. d Hence d = π) = π...7 solution The region bounded b the graph of = and the -ais over the interval [, ] consists of two right triangles, both above the ais. One triangle has area )) =, and the other has area )) = 9. Hence, InProblem Eercises.. and, refer to Figure. Let f) figure shown in the figure below. d = 9 + =. = f) )8) = )d FIGURE The two parts of the graph are semicircles. 6 Compute a) f) d b) f) d solution The region bounded b the graph 6 of = and the -ais over the interval [, ] is a. triangle Evaluate: above the a) ais f)d with base and b) height f)d. Consequentl, SOLUTIO. a) The region between the graph of f and -ais consists of a quarter of a circle of solution radius and Letaf)be quartergiven b Figure. )d = )) =. a) The definite integral of a circle of radius. Remembering that the definite integral computes signed area, we have f)dis the signed area of a semicircle of radius which lies below the -ais. Therefore, f) d = π) π) = π. f ) d = π ) = π b) For this integral, we compute the unsigned area between the. graph of f and the -ais, which consists of a quarter circle b) The definite integral or radius and a semicircles of radius. 6 f)dis the signed area of a semicircle of radius which lies below the -ais 6 and a semicircle of radius which lies f) above d = the -ais. Therefore, π) + π) = 9 π )d 6 f ) d = π ) π ) = π. solution The region bounded b the graph of = + and the -ais over the interval [, ] consists of a triangle below the ais with base and height, a triangle above the ais of base and height 6. and a Evaluate: triangle below a) the f)d ais of base b) and height f ). d Consequentl, solution Let f)be given b Figure.

5 Problem.. Determine the sign of the integral without calculating it. Draw a graph if necessar. d SOLUTIO. B smmetr, the positive area from the interval [, ] is cancelled b the negative area from [, ]. With the interval [, ] contributing more negative area, the definite integral must be negative... Problem..7 Calculate the integral: d. SOLUTIO. Since is negative on [, ] and positive on [, ], we have { ), =. Hence d = = ) d + [ ] = + ) d [ ] = =...7 Problem..78 Prove that.77 π/ cos d.6. SOLUTIO. cos is decreasing on the interval [, π/]. Hence, for π/, Since cosπ/) = /, Since cos).9,.77 π 8 π/ cos d cosπ/) cos cos). π/ = d π/ π/ cos d..9 d = π.9).6. 8 Therefore,.77 π cos

6 Problem..8 State whether true or false. If false, sketch the graph of a countereample. a) If f) >, then b f) d >. a b) If b f) d >, then f) >. a SOLUTIO. a) This is true in the case that b > a. If a > b, then b a f) d = a f) d and the integral b is negative. b) It is false that if b f) d >, then f) > for [a, b]. A countereample is f) = + a with a = and b =. We see that + )d = 7. >, et f ) = <. Here is the graph...8 6