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course MTH 173
11/5 22:45
Text Section 5.2
Problem 5.2.1
(a) The rectangle represent a right hand sum.
(b) Since the area of the rectangles is more than the area under the curve, thus the rectangles lead to an upper estimate.
(c) n is the number of rectangles which is 3
(d) Initial x value = 0, final x value = 6 and n = 3, thus delta x = (6-0)/3 = 2.
Problem 5.2.2
(a) The rectangle represent a left hand sum.
(b) Since the area of the rectangles is less than the area under the curve, thus the rectangles lead to an lower estimate.
(c) n is the number of rectangles which is 3
(d) Initial x value = 0, final x value = 6 and n = 3, thus delta x = (6-0)/3 = 2.
Problem 5.2.3
(a) The approximation is a left hand approximation, since the function is a decreasing function the right hand approximation will be less than the left hand approximation.
(b) The initial value of x is 0 thus a = 0 and the final value of x is 2 thus b = 2. n is the number of rectangles thus n = 6 and thus delta x = (b-a)/n = (2-0)/6 = 1/3 = 0.334 approx.
Problem 5.2.4
(a) When delta t = 4. t0 = 0, t1 = 4 and t2 = 8.
Thus f(t0) = 32, f(t1) = 24 and f(t2) = 0
Thus left hand sum = f(t0)*delta t + f(t1)*delta t = 32*4 + 24*4 = 224 square unit.
(b) When delta t = 4. t0 = 0, t1 = 4 and t2 = 8.
Thus f(t0) = 32, f(t1) = 24 and f(t2) = 0
Thus right hand sum = f(t1)*delta t + f(t2)*delta t = 24*4 + 0*4 = 96 square unit.
(c) When delta t = 2. t0 = 0, t1 = 2, t2 = 4, t3 = 6 and t4 = 8.
Thus f(t0) = 32, f(t1) = 30, f(t2) = 24, f(t3) = 14 and f(t4) = 0
Thus left hand sum = f(t0)*delta t + f(t1)*delta t + f(t2)*delta t + f(t3)*delta t
= 32*2 + 30*2 + 24*2 + 14*2 = 200 square unit.
(d) When delta t = 2. t0 = 0, t1 = 2, t2 = 4, t3 = 6 and t4 = 8.
Thus f(t0) = 32, f(t1) = 30, f(t2) = 24, f(t3) = 14 and f(t4) = 0
Thus right hand sum = f(t1)*delta t + f(t2)*delta t + f(t3)*delta t + f(t4)*delta t
= 30*2 + 24*2 + 14*2 + 0*2 = 136 square unit.
Problem 5.2.5
Given intgrl _1^4▒〖(x^2+x) dx〗 = ( x^3/3 + x^2/2 )41 = ( 64/3 + 16/2 - 1/3 - ½ ) = 21 + 7.5 = 28.5
Thus the value of the integral is 28.5.
Problem 5.2.8
Given to calculate the value of the integral of the function ln(y^2+1) for the y interval from y = 0 to y = 3. Using the calculator the value of the function will be approximated to 3.40585
Problem 5.2.11
Given that initial x = 0 and final x = 40. According to the values given I take n = 4
And thus delta x = 40/4 = 10.
From the given data the value of the left derivative will be
= 350*10 + 410*10 + 435*10 + 450*10 = 16,450
From the given data the value of the right derivative will be
= 410*10 + 435*10 + 450*10 + 460*10 = 17,550
The value of the integral will be between the left hand derivative which is 16,450 and the right hand derivative which is 17,550. The estimate value will be the average of the left hand and the right hand derivative which is 17,000.
Problem 5.2.14
To approximate intgrl _3^7▒〖1/(1+x) dx〗. Thus we use f(x) = 1 / (1+x) here we have a = 3 and b = 7 and n = 5.
Thus delta x = (b-a)/n = 4/5 = 0.8.
Thus x0 = 3, x1 = 3.8, x2 = 4.6, x3 = 5.4, x4 = 6.2 and x5 = 7
Thus the right hand sum will be = f(x1)*(delta x) + f(x2)*(delta x) + f(x3)*delta x + f(x4)*delta x + f(x5)*delta x.
Problem 5.2.17
The estimate of the integral will be the area under the curve for the graph. From x = -3 to x = 0 the integral is positive. The area under the curve for the positive values will thus depend upon the total number of squares under the positive part of the graph. The number of squares is approximately 3 full squares. Area of one square is 5 square unit. Thus area of 3 square is 15 square unit. Thus the integral of the function from x = -3 to x = 0 will be 15 square unit.
The area under the curve for the negative values will thus depend upon the total number of squares for x values from x = 0 to x = 5. Thus the number of square under the curve for x from x = 0 to x = 5 is 11 perfect approximate squares (formed by taking parts of square and putting with other part pf squares). The area of this 11 square is 55 square units and since this area is under x = 0 thus the area is negative and thus the area is -55 square unit.
Thus the value of the integral from x = -3 to x = 5 will be = 15 square unit - 55 square unit = -40 square unit.
Problem 5.2.20
The total area between y = 4 - x^2 and the x axis between x = 0 and x = 3.
The value of the total area between the function and the x axis will be the value of the integral provided that whenever the value of the integral is negative, the absolute value of the integral is to be considered.
For the interval of x = 0 and x = 3 the value of the function changes its sign that is the value becomes 0 when x = 2. Thus the integral of the function for the interval x = 0 to x = 2 is the be considered separately with the value of the integral for the interval x = 2 to x = 3.
Given function y = 4 - x^2.
The integral function will be f(x) = 4x - x^3/3
Thus the value of the integral for the interval x = 0 to x = 2 will be equal to f(2) - f(0).
f(2) = 8 - 8/3 and f(0) = 0. Thus the value of the integral will be 8 - 8/3
For the interval x = 2 to x = 3, the value of the integral will be f(3) - f(2)
Thus the integral value is = 12 - 9 - 8 + 8/3 = 8/3 - 5 which is negative
Thus the area under the curve for the interval x = 2 to x = 3 will be 5 - 8/3.
Thus the total area between the function y = 4 - x^2 and x axis will be 8 - 8/3 + 5 - 8/3
= 13 - 16/3 square units.
= 7.667 approx square units
Problem 5.2.23
The area of the function y = cost for t between 0 and pi/2 between the curve and the horizontal axis. The area will be the value of the integral between t = 0 and t = pi/2.
Given function y = cos(t)
The integral function thus will be f(t) = sin(t) and the value of the integral between t = 0 and pi/2 will be f(pi/2) - f(0) = sin(pi/2) - sin(0) = 1
Thus the value of the integral function y = cos(t) between t = 0 and t = pi/2 that is the area under curve for the same t interval is 1 square unit
Problem 5.2.24
The area of the function y = ln(x) between x = 1 and x = 4 will be nothing but the value of the integral of the function between the x values x = 1 and x = 4
The integral function will be f(x) = x( ln(x) - 1 )
The value of the integral for the x interval x = 1 to x = 4 will be f(4) - f(1)
Thus the value of the integral = 4ln(4) - 4 - ln(1) + 1 = 4ln(4) - 3
Thus the area under the curve y = ln(x) and the horizontal axis between x = 1 to x = 4 will be
4ln(4) - 3
Problem 5.2.26
Given function is y = cos(sqrt(x)) between x = 0 and x = 2. The area will be same as the value of the integral from x = 0 and x = 2, this is because the values of the function y between x values 0 to 2 is positive. Thus the integral function will be f(x) = 2*((sqrt(x)*sin (sqrt(x)) + cos(sqrt(x))).
Thus the value of the integral from x = 0 to x = 2 will be f(2) - f(0).
f(0) = 2*(0 + 1) = 2
f(2) = 2*((sqrt(2)sin(sqrt(2)) + cos(sqrt(2)))
Thus the area under the curve from x = 0 to x = 2 will be 2*((sqrt(2)sin(sqrt(2)) + cos(sqrt(2))) - 2
Problem 5.2.29
(a) The value of the integral will be the area under the curve from x = a to x = b which will be 13
(b) The value of the integral will be the area under the curve from x = b to x = c which is -2.
(c) The value of the integral will be the net sum of the values in part (a) and (b) which is 11.
(d) Here the mod function is used which basically indicates that no value will be negative and thus the negative integral will become positive. That is the part of the function from x = b to x = c will now be a mirror image along x axis as compared to the original graph. And thus the net value of the integral will be 13 + 2 = 15.
Problem 5.2.33
(a) The graph of the function f(x) = x(x+2)(x-1) will be a cubic curve tending to - infinity when x tends to - infinity and tends to + infinity when x tends to + infinity. The function has negative values for x values less than -2 and between 0 and 1. For the remaining x values the value is positive and the value is 0 for x = 0, x = -2 and x = 1.
(b) The total area between the curve and the x axis is divided in 2 parts, that is the area between the curve and x axis for x values between x = -2 to x = 0 and the area for the interval x = 0 to x = 1.
For the interval of x = -2 to x = 0 the area between the curve and the x axis is 8/3 square units and for the interval of x = 0 to x = 1 the area between the curve and the x axis is 5/12 square units
(c) The value of the integral will basically be the net area of the curve with the x axis provided the area’s ae considered along with their signs. We notice that for the interval x = 0 to x = 1 the area is negative and thus the integral for the interval x = 0 to x = 1 is -5/12. The value of the integral for the interval for x = -2 to x = 0 will be the area under the curve which is 8/3. Thus the net value of the integral for x value from x = -2 to x = 1 will be 9/4 = 2.25.
Problem 5.2.34
To calculate the integral of cos(sqrt(x)) from x = 0 to x = 4. The function however changes its sign from positive to negative at x = 2.4674 thus the function will have positive integral value for x value from x = 0 to x = 2.4674 and will have negative integral value from x = 2.4674 to x = 4. Thus in terms of area the value of the integral will be the net area under the curve and x axis provided that the sign of the area is considered.
The integral function is f(x) = 2*((sqrt(x)*sin (sqrt(x)) + cos(sqrt(x))).
The value of the integral will be f(4) - f(0).
f(4) = 2*(2sin(2) + cos(2)) and f(0) = 2 and thus the value of the integral will be
= 4sin(2) + 2cos(2) - 2.
Thus the value of the definite integral for the function will be 4sin(2) + 2cos(2) - 2.
Text section 5.3
Problem 5.3.3
Given that f(x) is measured in pounds and x is measured in feet, thus the units of the integral will be foot*pound.
Problem 5.3.6
The meaning of the integral will be the actual change in the population in billions from 2005 year to 2011. Thus the unit of the integral function will be people in billions.
Problem 5.3.10
We know that f(t) = F’(t)
It is given that F(t) = 3*(t^2) + 4t, a = 2 and b = 5.
Using the fundamental theorem of calculus the value of the integral function between the limits x = a and x = b will be F(b) - F(a).
F(b) = F(5) = 3*5*5 + 4*5 = 75 + 20 = 95.
F(a) = F(2) = 3*2*2 + 4*2 = 12 + 8 = 20.
Thus the value of the integral will be 95 - 20 = 75.
Problem 5.3.14
We know that f(t) = F’(t)
It is given that F(t) = tan(t), a = 0 and b = pi.
Using the fundamental theorem of calculus the value of the integral function between the limits x = a and x = b will be F(b) - F(a).
F(b) = F(pi) = tan(pi) = 0.
F(a) = F(0) = tan(0) = 0.
Thus the value of the integral will be 0 - 0 = 0.
Problem 5.3.17
(a) Given that F(t) = (1/2)*(sin^2(t))
Thus F’(t) will be sin(t)*cos(t)
(b) To find integral for the function f(t) = sin(t)*cos(t) for the interval of t = 0.2 to t = 0.4
(i)
Using a calculator I figured out the integral to be equal to an approximate value of 0.056
(ii) Using the fundamental theorem of calculus we can calculate the value of the integral.
We know that F’(t) = f’(t) = F(t)
Thus the integral of the function f’(t) within the interval x = 0.2 to x = 0.4 will be nothing but
F(0.4) - F(0.2) = (1/2)*sin^2(0.4) - (1/2)*sin^2(0.2) = 0.05608857
Problem 5.3.20
Given that the rate function is r = f(t) where r is in gallons per minute and t is in minutes. If F(t) is the function of volume of oil in tanker than F’(t) = f(t). Thus the integral function of the f(t) function will be F(t) and thus the value of the integral for the first hour which also represents the change in the oil that leaks out of the tanker in the first hour will be equal to F(1) - F(0). Thus the definite expression for the same will be f(t)dt integrated over the limits b = 1 and a = 0.
Problem 5.3.22
(a) To estimate the total number of acres defaced in extracting the next 4 million tons of coal we draw 4 rectangles under the curve. Using that basically we calculate left hand sum. We have x0 = 0, x1 = 1, x2 = 2, x3 = 3, x4 = 4 and x5 = 5. We have f(t0) = 0.2, f(t1) = 0.4, f(t2) = 1, f(t3) = 2, f(t4) = 3.4.
Thus the left hand sum will be f(t0)*1 + f(t1)*1 + f(t2)*1 + f(t3)*1 = 3.6 acres.
(b) Re-estimating the number of acres defaced using rectangles above the curve will be
= f(t1)*1 + f(t2)*1 + f(t3)*1 + f(t4)*1 = 6.8 acres
(c) We have the right hand and the left hand estimate of the number of acres defaced, thus the actual value of the area defaced will lie between this value and an estimate of the same would be an average of the 2 values. Thus the estimate of the actual value of the number of acres of land defaced will be 5.2.
Problem 5.3.27
Given function for the amount of waste a company produces in tons per week is approximated by
W = 3.75*(e^-0.008t) where t is weeks since January 1, 2005. In the year 2005 we have 52 weeks.
The cost incurred to the company is $15/ton.
The anti-derivative of the function is w(t) = (-3.75/0.008)*(e^-0.008t)
Thus the total waste accumulated during the entire year 2005 will be integral of W(t) function from t = 0 to t = 52 weeks which by fundamental theorem of calculus will be w(52) - w(0)
w(0) = -468.75
w(52) = -309.2251
thus w(52) - w(0) = -309.2251 + 468.75 = 159.5249
thus the total cost the company pays = 159.5249 tons * $15/ton = $2392.8735
Problem 5.3.32
Given that the rate at which water is pumped out of a holding tank is r(t) = 5 - 5e^(-0.12t) liters/min where t is time in min since the pump was started.
The anti-derivative function of r(t) will be R(t) = 5t + (5/0.12)*(e^(-0.12t))
Thus we have R’(t) = r(t)
The amount of water that is pumped out during 1 hours will be the value of the integral of the r(t) with a = 0 and b = 60.
Using the fundamental theorem of calculus we know that the integral of the function r(t) from a = 0 to b = 60 is same as R(60) - R(0).
R(0) = 0 + (5/0.12) = 41.6666667 approx
R(60) = 300 + (5/0.12)*(e^-7.2) = 300.0311077 approx
Thus the value of the integral will be 258.364441
The initial amount of water in the tank is 1000 liters, thus the amount of water after 1 hours is 1000 - 258.364441 liters = 741.635559 liters.
Problem 5.3.34
The given graph is of f’(x) function.
The value of (f(4)-f(2))/2, f(3) - f(2), f(4) - f(3) are to be arranged in increasing order.
f(x) is the integral function and using the fundamental theorem of calculus we know that f(4) - f(2), f(3) - f(2) and f(4) - f(3) represent the area under the curve for x intervals from 2 to 4, 2 to 3 and 3 to 4 respectively considered along with its sign.
Interpreting the areas under the graph we can see that the absolute value of f(3) - f(2) is greater than the absolute value of (f(4) - f(2)) / 2 which is in turn greater than the absolute value of f(4) - f(3). But all this values are negative and when considered the signs the ascending order of the values will be
f(3) - f(2) < (f(4) - f(2))/2 < f(4) - f(3)
Problem 5.3.38
The 5 graphs given below are graphs of v vs. t where v is velocity and t is time. The integral of the curve will be the total distance covered. The derivative or the slope function will be the acceleration function.
(a) Has a constant acceleration - the slope of the v vs. t graph should be constant which is graph (V).
(b) Ends up farthest to the left of where it started - Since the integral that is area under curve with sign determines the distance the particle is from the starting position. Farthest on the left will be the graph which has the least value of the integral which is (IV)
(c) Ends up farthest from its starting point - The graph whose absolute value of the integral will be greatest will end up farthest from its starting point. It can be seen that the graphs (I), (II) and (IV) are graphs who have negative value of the integral as well as positive value and thus the net absolute value tend to decrease. Whereas for the other (III) and (V) graph we can see that the area the function will be very small for graph (V) as compared to graph (III) and thus graph (III) tend to have the maximum area under the curve as compared to all the other graphs. Thus the particle will end up farthest from the starting point in case of graph (III).
(d) Experiences the greatest initial acceleration - The graph which has the greatest initial slope will indicate the greatest initial acceleration. The graph (II) indicates this.
(e) Has the greatest average velocity - average velocity will be the sum of all the velocities divided by the number of velocities considered, or that would also be equal to the difference between the initial and the final position of the particle divided by the time taken to change that distance. Since the initial distance and the time taken is same for all the graphs the only thing that will now matter is the final distance. Since graph (III) represents the greatest distance therefore the change in distance will be greatest for graph (III) and thus eventually the average velocity will be greatest for (III).
(f) Has the greatest average acceleration - Average acceleration is defined as the change in velocity / time. That is (final velocity - initial velocity)/(time taken). The difference in the velocity is greatest in graph (I), and time taken is same for all graphs. Thus (I) represents the greatest average acceleration.
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