Calculus Questions

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1 0 1 Points DETAILS A T Find the area of the region that lies inside the first curve and outside the second curve r 3 cos 8 r 1 cos 8 PREVIOUS ANSWERS x Enhanced Feedback Please try again keeping in mind that you should first find the intersection points Then use the formula for the area between two parametric curves which is A Watch it Ion con dl where 2
Calculus
Definite Integrals
1 0 1 Points DETAILS A T Find the area of the region that lies inside the first curve and outside the second curve r 3 cos 8 r 1 cos 8 PREVIOUS ANSWERS x Enhanced Feedback Please try again keeping in mind that you should first find the intersection points Then use the formula for the area between two parametric curves which is A Watch it Ion con dl where 2
8 7 1 70 1 n Lan n 1 As 4 1 this means that p series converges 09 7 07 Step 4 We have found that the given power series converges for 1x1 7 and also for the endpoint x 7 Finally we must test the second endpoint of the interval e 8 1 n converges Therefore x 7 in the interval of convergence n 1 7 1 By the Alternating Series Test this series Select To conclude find the interval 1 of convergence of the series Enter your answer using interval notation back
Calculus
Application of derivatives
8 7 1 70 1 n Lan n 1 As 4 1 this means that p series converges 09 7 07 Step 4 We have found that the given power series converges for 1x1 7 and also for the endpoint x 7 Finally we must test the second endpoint of the interval e 8 1 n converges Therefore x 7 in the interval of convergence n 1 7 1 By the Alternating Series Test this series Select To conclude find the interval 1 of convergence of the series Enter your answer using interval notation back
DETAILS b sin br 114 0 1 Points Find the solution of the differential equation that satisfies the given initial condition dA Ab cos br A 0 6 dr Need Help x PREVIOUS Read It
Calculus
Definite Integrals
DETAILS b sin br 114 0 1 Points Find the solution of the differential equation that satisfies the given initial condition dA Ab cos br A 0 6 dr Need Help x PREVIOUS Read It
Therefore k 30 In 2 30 Step 4 Now remembering that In x n In x and that eln z z we have y t 50e t 30 In 2 mg
Calculus
Application of derivatives
Therefore k 30 In 2 30 Step 4 Now remembering that In x n In x and that eln z z we have y t 50e t 30 In 2 mg
P 3 0 1 Points Find the solution of the differential equation that satisfies the given initial condition dp dt DETAILS 7 3 ad Help 7 Pt P 1 3 3 PREVIOUS 7 3 Read It X
Calculus
Application of derivatives
P 3 0 1 Points Find the solution of the differential equation that satisfies the given initial condition dp dt DETAILS 7 3 ad Help 7 Pt P 1 3 3 PREVIOUS 7 3 Read It X
DETAILS 9 0 1 Points PREVIOUS ANSW Find the area of the region that lies inside the first curve and outside the second curve r 16 sin 8 11 71 x r 8
Calculus
Application of derivatives
DETAILS 9 0 1 Points PREVIOUS ANSW Find the area of the region that lies inside the first curve and outside the second curve r 16 sin 8 11 71 x r 8
Sketch the region in the plane consisting of points whose polar coordinates satisfy the given conditions r 7 0 0 8 6 4 2 y 8 6 4 2 2 4 6 8F y 8 6 4 2 2 4 6 8 X 8 8 6 4 2 4 6 y 8 6 2 4 2 2 4 6 8F y 8 6 2 2 2 4 4 6 6 8 X 8
Calculus
Definite Integrals
Sketch the region in the plane consisting of points whose polar coordinates satisfy the given conditions r 7 0 0 8 6 4 2 y 8 6 4 2 2 4 6 8F y 8 6 4 2 2 4 6 8 X 8 8 6 4 2 4 6 y 8 6 2 4 2 2 4 6 8F y 8 6 2 2 2 4 4 6 6 8 X 8
2 O T 2 2 N 2 O DO T N T N 2 b 4 2
Calculus
Application of derivatives
2 O T 2 2 N 2 O DO T N T N 2 b 4 2
DETAILS 11 0 1 Points Find the area of the region that lies inside the first curve and outside the second curve r 3 cos 8 r 1 cos 8 A T x Enhanced Feedback Please try again keeping in mind that you should first find the intersection points Then use the formula for the area between two parametric curves which is A toll
Calculus
Definite Integrals
DETAILS 11 0 1 Points Find the area of the region that lies inside the first curve and outside the second curve r 3 cos 8 r 1 cos 8 A T x Enhanced Feedback Please try again keeping in mind that you should first find the intersection points Then use the formula for the area between two parametric curves which is A toll
W S 0 01 n 1 Step 4 Recall the following theorem 8 n 1 For the sum 0 01 n 1 we have a 0 01 The geometric seriesar 1 a ar ar a ar ar is convergent if Irl 1 and its sum is n 1 7 1 0 01 n 1 0 81 81 n 1 1 10 01 0 01 100 99 0 01 100 99 the geometric series ar 1 n 1 Step 5 We have found the value of the geometric sequence with common ratio 0 01 Therefore we can now wice the repeating fraction 0 81 as a ratio of integers as follows 0 01 1 81 100 100 99 17 Inci 1 and r 0 01 1 Therefore we have the following
Calculus
Limits & Continuity
W S 0 01 n 1 Step 4 Recall the following theorem 8 n 1 For the sum 0 01 n 1 we have a 0 01 The geometric seriesar 1 a ar ar a ar ar is convergent if Irl 1 and its sum is n 1 7 1 0 01 n 1 0 81 81 n 1 1 10 01 0 01 100 99 0 01 100 99 the geometric series ar 1 n 1 Step 5 We have found the value of the geometric sequence with common ratio 0 01 Therefore we can now wice the repeating fraction 0 81 as a ratio of integers as follows 0 01 1 81 100 100 99 17 Inci 1 and r 0 01 1 Therefore we have the following
Since the half life is 30 years then y 30 25 Step 3 Thus 25 50e30k or 30k Therefore k H 25 mg
Calculus
Application of derivatives
Since the half life is 30 years then y 30 25 Step 3 Thus 25 50e30k or 30k Therefore k H 25 mg
This question has several parts that able to come back to the skipped part The half life of cesium 137 is 30 years Suppose we have a 50 mg sample Exercise a Find the mass that remains after t years Step 1 Let y t be the mass in mg remaining after t years Then we know the following y t y 0 ekt ekt
Calculus
Application of derivatives
This question has several parts that able to come back to the skipped part The half life of cesium 137 is 30 years Suppose we have a 50 mg sample Exercise a Find the mass that remains after t years Step 1 Let y t be the mass in mg remaining after t years Then we know the following y t y 0 ekt ekt
Solve for x cos x cos 2x 0 0 2n X Select 3 correct answer s KO 4 60 03 Gla co No solution 02 2 W H 560 n
Calculus
Application of derivatives
Solve for x cos x cos 2x 0 0 2n X Select 3 correct answer s KO 4 60 03 Gla co No solution 02 2 W H 560 n
Solve for x 3 cos x 1 cos x over the interval 0 2 If needed give your answer s as an approximation rounded to the nearest 2 decimal places If there is more than one answer separate your answers using a comma with no spaces between symbols Ex 5 34 3 20
Calculus
Differential equations
Solve for x 3 cos x 1 cos x over the interval 0 2 If needed give your answer s as an approximation rounded to the nearest 2 decimal places If there is more than one answer separate your answers using a comma with no spaces between symbols Ex 5 34 3 20
When solving the following equation using adjusted interval methods cos 0 over the interval 0 4 what is the ADJUSTED INTERVAL that you would use to solve 0 8T None of these 0 2x 0 4 0
Calculus
Differentiation
When solving the following equation using adjusted interval methods cos 0 over the interval 0 4 what is the ADJUSTED INTERVAL that you would use to solve 0 8T None of these 0 2x 0 4 0
PREVIOUS ANSWERS the area of the region that is bounded by the given curve and lies in the specified sector r e 3 t 4 Os 3128 2 DETAILS ints 12280 33 X
Calculus
Application of derivatives
PREVIOUS ANSWERS the area of the region that is bounded by the given curve and lies in the specified sector r e 3 t 4 Os 3128 2 DETAILS ints 12280 33 X
1 0 1 Points DETAILS Find the area of the region that is bounded by the given curve and lies in the specified sector r 180 0 8 I 2 3 x T
Calculus
Application of derivatives
1 0 1 Points DETAILS Find the area of the region that is bounded by the given curve and lies in the specified sector r 180 0 8 I 2 3 x T
Answer Within three units of the xz plane
Calculus
Application of derivatives
Answer Within three units of the xz plane
Solve for x cos x cos 2x 0 X 0 2
Calculus
Differential equations
Solve for x cos x cos 2x 0 X 0 2
X Select 4 correct answer s 4 3 00 11 57 12 03 017 No solution 12 13m 12 2 5 6
Calculus
Application of derivatives
X Select 4 correct answer s 4 3 00 11 57 12 03 017 No solution 12 13m 12 2 5 6
Solve for x 2 sin x 2 cos x 0 2 X
Calculus
Vector Calculus
Solve for x 2 sin x 2 cos x 0 2 X
Choose ALL of the following that are equivalent to a Pythagorean identity Select 6 correct answer s O cot x csc x 1 Otan x sec x 1 csc a cot x 1 tan x sec x 1 Osin x 1 cos x Ocot x csc x 1 Osec x tan x 1 sin x 1 cos x cos r sin x 1 Ocot x csc x 1 O sin x cos x 1 Osec x 1 tan x K
Calculus
Limits & Continuity
Choose ALL of the following that are equivalent to a Pythagorean identity Select 6 correct answer s O cot x csc x 1 Otan x sec x 1 csc a cot x 1 tan x sec x 1 Osin x 1 cos x Ocot x csc x 1 Osec x tan x 1 sin x 1 cos x cos r sin x 1 Ocot x csc x 1 O sin x cos x 1 Osec x 1 tan x K
Given the identity cos x sin x 1 select all of the following that are equivalent to it
Calculus
Limits & Continuity
Given the identity cos x sin x 1 select all of the following that are equivalent to it
Determine which of the following identity replacements is the best accurate choice in solving the equation 5 cos x sin x 1 5 1 sin x sin x 1 05 1 2 sin r sin x 1 05 1 sin x sin x 1 None of these O5 sin 2 sin x 1 D
Calculus
Limits & Continuity
Determine which of the following identity replacements is the best accurate choice in solving the equation 5 cos x sin x 1 5 1 sin x sin x 1 05 1 2 sin r sin x 1 05 1 sin x sin x 1 None of these O5 sin 2 sin x 1 D
Determine which of the following double identity replacements is the best accurate choice in solving the equation cos x cos 2x 0 cos x 2 cos x sin x 0 None of these O cos x cos r sin x 0 cos x 2 cos x 1 0 cos x 1 2 sin x 0 G
Calculus
Application of derivatives
Determine which of the following double identity replacements is the best accurate choice in solving the equation cos x cos 2x 0 cos x 2 cos x sin x 0 None of these O cos x cos r sin x 0 cos x 2 cos x 1 0 cos x 1 2 sin x 0 G
Find the divergence of the field F 5ze4xyzi 5ze i 5ze div F j 5x4xyzk
Calculus
Application of derivatives
Find the divergence of the field F 5ze4xyzi 5ze i 5ze div F j 5x4xyzk
1 2 Points DETAILS PREVIOUS ANSWERS SCALCET9 11 10 021 Find the Taylor series for f x centered at the given value of a Assume that f has a power series expansion Do not show that R x 0 x5 3x x a 4 f x 1220 1425 x 2 658 x 2 R Find the associated radius of convergence R 2 323 3 x 3 3 20 x a 4 x 55 0 X
Calculus
Limits & Continuity
1 2 Points DETAILS PREVIOUS ANSWERS SCALCET9 11 10 021 Find the Taylor series for f x centered at the given value of a Assume that f has a power series expansion Do not show that R x 0 x5 3x x a 4 f x 1220 1425 x 2 658 x 2 R Find the associated radius of convergence R 2 323 3 x 3 3 20 x a 4 x 55 0 X
Use the divergence theorem to find the outward flux of F across the boundary of the region D F 2y x i z y j 2y 2x k D The cube bounded by the planes x 3 y 3 and z 3 The outward flux is Type an exact answor
Calculus
Differential equations
Use the divergence theorem to find the outward flux of F across the boundary of the region D F 2y x i z y j 2y 2x k D The cube bounded by the planes x 3 y 3 and z 3 The outward flux is Type an exact answor
lim 318 n 1 an lim 816 lim 818 x1 7 n 1 450 1 X 4 n n lim n 1 a n 1 lim 818 an 4 n n 1 4 Step 2 We have found the following limit from the Ratio Test x 7 n n 1 Submit Skip you cannot come back 71 By the Ratio Test the limit converges if Q n x FIX 1 and diverges if 1 Therefore the limit converges for Ixl if 21 The limit diverges for x Again recall that the radius of convergence is the positive value R such that the power series in this case centered at 0 converges if x R and diverges if x R Therefore the radius of convergence for the given power series is R
Calculus
Limits & Continuity
lim 318 n 1 an lim 816 lim 818 x1 7 n 1 450 1 X 4 n n lim n 1 a n 1 lim 818 an 4 n n 1 4 Step 2 We have found the following limit from the Ratio Test x 7 n n 1 Submit Skip you cannot come back 71 By the Ratio Test the limit converges if Q n x FIX 1 and diverges if 1 Therefore the limit converges for Ixl if 21 The limit diverges for x Again recall that the radius of convergence is the positive value R such that the power series in this case centered at 0 converges if x R and diverges if x R Therefore the radius of convergence for the given power series is R
27 5 Points DETAILS Test the series for convergence or divergence 2 2 4 6 8 S 6 7 SCALCET9 Evaluate the following limit lim bn 318 Identify b Assume the series starts at n 1 0 10 8 Since lim b 0 and bo 1 810 4 b for all n the series converges
Calculus
Limits & Continuity
27 5 Points DETAILS Test the series for convergence or divergence 2 2 4 6 8 S 6 7 SCALCET9 Evaluate the following limit lim bn 318 Identify b Assume the series starts at n 1 0 10 8 Since lim b 0 and bo 1 810 4 b for all n the series converges
Find the curl of the vector field F F x y z i 9x y 7z j 6x 8y z k curl F i j k
Calculus
Application of derivatives
Find the curl of the vector field F F x y z i 9x y 7z j 6x 8y z k curl F i j k
Find the curl of the vector field F F xy z i yz x j xz y k curl F i i k
Calculus
Vector Calculus
Find the curl of the vector field F F xy z i yz x j xz y k curl F i i k
10 1 2 Points DETAILS Determine whether the geometric series is convergent or divergent n 1 convergent divergent 16 0 64 1 PREVIOUS ANSWERS SCALCET9 11 2 027 x If it is convergent find its sum If the quantity diverges enter DIVERGES 44 4444
Calculus
Indefinite Integration
10 1 2 Points DETAILS Determine whether the geometric series is convergent or divergent n 1 convergent divergent 16 0 64 1 PREVIOUS ANSWERS SCALCET9 11 2 027 x If it is convergent find its sum If the quantity diverges enter DIVERGES 44 4444
Find the divergence of the field F 4x 6y 5z i 8x 6y 7z j 5x 6y 3z k AMAN neuen Einfl Wirtu duben
Calculus
Application of derivatives
Find the divergence of the field F 4x 6y 5z i 8x 6y 7z j 5x 6y 3z k AMAN neuen Einfl Wirtu duben
f x n 1 1 2 Points Find the Maclaurin series for f x using the definition of a Maclaurin series Assume that f has a power series expansion Do not show that R x 0 f x In 1 5x DETAILS R 15 PREVIOUS ANSWERS 1 Find the associated radius of convergence R n 1 x 1 5 1 n X SCALCE19
Calculus
Application of derivatives
f x n 1 1 2 Points Find the Maclaurin series for f x using the definition of a Maclaurin series Assume that f has a power series expansion Do not show that R x 0 f x In 1 5x DETAILS R 15 PREVIOUS ANSWERS 1 Find the associated radius of convergence R n 1 x 1 5 1 n X SCALCE19
Step 5 Now we must find the radius of convergence for the Maclaurin series First we set up the ratio test as follows a lim n 1 318 an lim n48 n 2 816 n 1 x lim Step 7 Step 6 Now we can simplify and evaluate the limit 4 n 2 x 1 4 n 1 x 1 Submit Skip you cannot come back 4 n 1 x By the Ratio Test the series converges when lim 816 an 1 an 4 n 1 x That is the positive number R such that the series converges if x al R and diverges if Ix al R We can use the ratio test to find R n 0 x 1 Therefore the radius of convergence is R 1 X X
Calculus
Application of derivatives
Step 5 Now we must find the radius of convergence for the Maclaurin series First we set up the ratio test as follows a lim n 1 318 an lim n48 n 2 816 n 1 x lim Step 7 Step 6 Now we can simplify and evaluate the limit 4 n 2 x 1 4 n 1 x 1 Submit Skip you cannot come back 4 n 1 x By the Ratio Test the series converges when lim 816 an 1 an 4 n 1 x That is the positive number R such that the series converges if x al R and diverges if Ix al R We can use the ratio test to find R n 0 x 1 Therefore the radius of convergence is R 1 X X
Consider the following series 1 2 4 4 n 1 Does the function f x O Yes O No 1 x 4 satisfy the conditions of the Integral Test Evaluate the following integral If it is convergent evaluate it If the quantity diverges enter DIVERGES 1 x 2 40x dx 4 Orient finite the series is Select
Calculus
Indefinite Integration
Consider the following series 1 2 4 4 n 1 Does the function f x O Yes O No 1 x 4 satisfy the conditions of the Integral Test Evaluate the following integral If it is convergent evaluate it If the quantity diverges enter DIVERGES 1 x 2 40x dx 4 Orient finite the series is Select
Suppose an and b are series with positive terms and b is known to be divergent a If an bn for all n what can you say about an Why a converges if and only if n an 2 b a converges if and only if 2a 2b a diverges by the Comparison Test a converges by the Comparison Test We cannot say anything about an b If an b for all n what can you say about an Why O We cannot say anything about an S a converges if and only if an 4 a diverges by the Comparison Test a converges if and only if an an n converges by the Comparison Test
Calculus
Application of derivatives
Suppose an and b are series with positive terms and b is known to be divergent a If an bn for all n what can you say about an Why a converges if and only if n an 2 b a converges if and only if 2a 2b a diverges by the Comparison Test a converges by the Comparison Test We cannot say anything about an b If an b for all n what can you say about an Why O We cannot say anything about an S a converges if and only if an 4 a diverges by the Comparison Test a converges if and only if an an n converges by the Comparison Test
0 01 0 01 0 0001 0 01 2 Continuing this pattern we see that 1 0 01 0 0001 can be expressed as the geometric series 00 0 01 n 1 Step 4 Recall the following theorem For the sum n 1 00 0 01 The geometric series n 1 n 1 0 01 1 1 0 01 1 1 ar 1 a ar ar is convergent if r 1 and its sum is we have a 1 Submit Skip you cannot come back n 1 ar 1 a and r 0 01 1 Therefore we have the following 1 r r 1
Calculus
Application of derivatives
0 01 0 01 0 0001 0 01 2 Continuing this pattern we see that 1 0 01 0 0001 can be expressed as the geometric series 00 0 01 n 1 Step 4 Recall the following theorem For the sum n 1 00 0 01 The geometric series n 1 n 1 0 01 1 1 0 01 1 1 ar 1 a ar ar is convergent if r 1 and its sum is we have a 1 Submit Skip you cannot come back n 1 ar 1 a and r 0 01 1 Therefore we have the following 1 r r 1
Supposea and b bare series with positive terms and b is known to be convergent n a If a b for all n what can you say about a Why Oa diverges by the Comparison Test a converges if and only if a 2b a converges if and only if a 4b a converges by the Comparison Test an O O We cannot say anything about an b If an b for all n what can you say about an Why O We cannot say anything about an b a converges if and only if n sans br an 2 a converges by the Comparison Test a diverges by the Comparison Test bn converges if and only if n sans br 4 Ox an
Calculus
Application of derivatives
Supposea and b bare series with positive terms and b is known to be convergent n a If a b for all n what can you say about a Why Oa diverges by the Comparison Test a converges if and only if a 2b a converges if and only if a 4b a converges by the Comparison Test an O O We cannot say anything about an b If an b for all n what can you say about an Why O We cannot say anything about an b a converges if and only if n sans br an 2 a converges by the Comparison Test a diverges by the Comparison Test bn converges if and only if n sans br 4 Ox an
9 0 1 Points DETAILS 8 7 PREVIOUS ANSWERS Determine whether the geometric series is convergent or divergent If it is convergent find its sum If the quantity diverges enter DIVERGES 49 343 8 64 SCALCET9 11 2 024 MY NO
Calculus
Limits & Continuity
9 0 1 Points DETAILS 8 7 PREVIOUS ANSWERS Determine whether the geometric series is convergent or divergent If it is convergent find its sum If the quantity diverges enter DIVERGES 49 343 8 64 SCALCET9 11 2 024 MY NO
Determine whether the series converges or diverges 1 n 8 n 8 n 1 converges O diverges Need Help Read It
Calculus
Application of derivatives
Determine whether the series converges or diverges 1 n 8 n 8 n 1 converges O diverges Need Help Read It
where a 0 00 n 1 b 1 1 and c 2 Step 3 We have noticed the following for the first terms of our sequence 1 0 01 0 01 0 01 0 0001 0 01 Continuing this pattern we see that 1 0 01 0 0001 can be expressed as the geometric series I
Calculus
Application of derivatives
where a 0 00 n 1 b 1 1 and c 2 Step 3 We have noticed the following for the first terms of our sequence 1 0 01 0 01 0 01 0 0001 0 01 Continuing this pattern we see that 1 0 01 0 0001 can be expressed as the geometric series I
0 000081 0 81 1 10000 0 01 Step 2 We have written the given repeating decimal as follows 0 81 81 1 0 01 0 0001 Notice that each term of the sequence in the parentheses can 1 0 01 0 01 b 0 0001 0 01 where a 0 01 b 0 0001 cannot come back and c be written written as a power of 0 01
Calculus
Vector Calculus
0 000081 0 81 1 10000 0 01 Step 2 We have written the given repeating decimal as follows 0 81 81 1 0 01 0 0001 Notice that each term of the sequence in the parentheses can 1 0 01 0 01 b 0 0001 0 01 where a 0 01 b 0 0001 cannot come back and c be written written as a power of 0 01
Determine whether the geometric series is convergent or divergent 80 16 0 64 1 n 1 O convergent O divergent If it is convergent find its sum If the quantity diverges enter DIVERGES
Calculus
Application of derivatives
Determine whether the geometric series is convergent or divergent 80 16 0 64 1 n 1 O convergent O divergent If it is convergent find its sum If the quantity diverges enter DIVERGES
Calculate the sum of the series Sn n 1 3n 1 00 n 1 an whose partial sums are given
Calculus
Limits & Continuity
Calculate the sum of the series Sn n 1 3n 1 00 n 1 an whose partial sums are given
Tutorial Exercise Express the number as a ratio of integers 0 81 0 81818181 Step 1 To write 0 81 as a ratio of integers first write it in expanded form 0 81 as 0 818181 0 81 0 0081 0 000081 Factoring out 0 81 from each of these terms gives the following 0 81 0 81 0 0081 1 0 81 1 0 000081 0 81
Calculus
Definite Integrals
Tutorial Exercise Express the number as a ratio of integers 0 81 0 81818181 Step 1 To write 0 81 as a ratio of integers first write it in expanded form 0 81 as 0 818181 0 81 0 0081 0 000081 Factoring out 0 81 from each of these terms gives the following 0 81 0 81 0 0081 1 0 81 1 0 000081 0 81
a What is the difference between a sequence and a series b A sequence is an ordered list of numbers whereas a series is an unordered list of numbers A sequence is an ordered list of numbers whereas a series is the sum of a list of numbers A sequence is the sum of a list of numbers whereas a series is an ordered list of numbers A sequence is the sum of a list of numbers whereas a series is an unordered list of numbers A sequence is an unordered list of numbers whereas a series is the sum of a list of numbers What is a convergent series What is a divergent series A series is convergent if the sequence of Select is a Select sequence A series Select if it is not convergen
Calculus
Limits & Continuity
a What is the difference between a sequence and a series b A sequence is an ordered list of numbers whereas a series is an unordered list of numbers A sequence is an ordered list of numbers whereas a series is the sum of a list of numbers A sequence is the sum of a list of numbers whereas a series is an ordered list of numbers A sequence is the sum of a list of numbers whereas a series is an unordered list of numbers A sequence is an unordered list of numbers whereas a series is the sum of a list of numbers What is a convergent series What is a divergent series A series is convergent if the sequence of Select is a Select sequence A series Select if it is not convergen
Find the area of the region that lies inside the first curve and outside the second curve r 16 sin 8 r 8
Calculus
Application of derivatives
Find the area of the region that lies inside the first curve and outside the second curve r 16 sin 8 r 8
Use the Integral Test to determine whether the series is convergent or divergent 00 n 1 76 Evaluate the following integral X 6dx Since the integral Select finite the series is Select
Calculus
Application of derivatives
Use the Integral Test to determine whether the series is convergent or divergent 00 n 1 76 Evaluate the following integral X 6dx Since the integral Select finite the series is Select