devotion & mathematics

                            Line Integrals                  Introduction                        Problems & Solutions                                                          Back

             Integration over Rectangular Regions                                                                Integration over Non-Rectangular Bounded Regions                                                                                                     Repeated integrals                                                                               ...

                                Double Integrals                                                      Problems & Solutions                                                               Back

                                  Integral Calculus        Historical Background of Integral Calculus                                                                      Line Integrals                                                                                           Double Integrals                                

                                                  Comparison Test                  Proof                                            Problems & Solutions                                                                                       Back

Comparison Test : Problems & Solutions               Test for convergence of Σ 1/(n 2 +n)                                                                            Back Test for convergence of Σ 1/ logn Test for convergence of Σ n 3 /(n 5 +4n 4 +7) Test for convergence of Σ 1/n! Test for convergence of Σ log(1/n) Test for convergence of Σ logn/(2n 3 -1)  #comparison #test #problems #& #solutions #infinite #series

Comparision Test -1 :   Statement : If Σu n and Σv n are two series of non-negative terms such that                                Back a)                                           There is a positive integer m and k 𝞊 R +  , 0 ≤  u n ≤ k v n   ∀ n ≥ m and                       b)       Σv n   is convergent then Σu n is convergent.      Proof : Comparison test -2 : Statement  : If Σu n  and Σv n  are two series of non-negative terms such that  a)                                          There is a positive integer m and k  𝞊  R +  ...

  P- test for series  : Problems & Solutions                                                                                                                               Back * Test for convergence of Σ 1/n 2 *Test for convergence of Σ 1/n 3/2 .                                                             * Test for convergence of Σ 1/√n.                                     * Test for convergence of Σ 1/n 2 .     ...

                                                                                                                                    P-Test for series :  Statement & Proof                                                           Back                                                                               ...

                      P- TEST FOR SERIES                                                                       PROOF                                                        PROBLEMS &  SOLUTIONS                                                                                                                         ...

                     INFINITE SERIES                                               Definition   :                       If { u n }  is a sequence of real numbers and  S n = u 1 +u 2 +u 3 +…+u n ,for some +ve integer n, then the sequence  { s n } is called an infinite series.    The number u n is called the n th term of the series.    The number s n is called the n th partial sum of the series. The infinite series { s n } is denoted by Σ u n = u 1 +u 2 +u 3 +… Convergence of Series  :                                                            ...

                             INFINITE SERIES                                               Introduction                         P-Test                           Comparision test                          Limit Comparison Test                                                                      Cauchy's nth root test                                   ...

  PROBLEM :                       Using polar coordinates, show that ∫dx ∫√X 2 +y 2     dy =1/6[√2 + log ( 1+√2) ] SOLUTION :  Evaluate ∬ f(x,y) dxdy where f(x,y)= (2y-1)/ x+1 , and E is the region bounded by x=0,y=0, y=2x-        4.  *    Change the order of integration and hence show that ∬dx dy/[ ( 1+e^y) √(1-x^2-y^2)]   = (𝝿 /2)                    log(2e/(1+e) where x-0 to x=1 and y=0 to y=√(1-x^2)  *    In the integral ∬ (4-y) dydx, change the order of integration and evaluate the integral where x=2 to            x=4 and y=4/x to y=(20-4x)/(8-x)  *   Sketch the region of integration and evaluate ∬ xsiny dydx where x=0 to x=𝛑 and y-0 to y=x.  *   Sketch the region of integration and write an equivalent double integral with th...

PROBLEM :                          Evaluate ∬ e^(x 2 +y 2 ) dxdy , where E is the semi circular region bounded by the X-axis and                               the curve y= √ (1-x 2 ). SOLUTION :    Evaluate ∬ f(x,y) dxdy where f(x,y)= (2y-1)/ x+1 , and E is the region bounded by x=0,y=0, y=2x-        4.  *    Change the order of integration and hence show that ∬dx dy/[ ( 1+e^y) √(1-x^2-y^2)]   = (𝝿 /2)                    log(2e/(1+e) where x-0 to x=1 and y=0 to y=√(1-x^2)  *    In the integral ∬ (4-y) dydx, change the order of integration and evaluate the integral where x=2 to            x=4 and y=4/x to y=(20-4x)/(8-x)  *   Sketch the region of integration and evaluate ∬ xsi...

PROBLEM :                          Evaluate ∬ xy(x+y) 2 /(x 2 +y 2 ) dxdy where E is region bounded y=0,y=x,x 2 +y 2 =a 2                         in the first quadrant.  SOLUTION :  Evaluate ∬ f(x,y) dxdy where f(x,y)= (2y-1)/ x+1 , and E is the region bounded by x=0,y=0, y=2x-        4.  *    Change the order of integration and hence show that ∬dx dy/[ ( 1+e^y) √(1-x^2-y^2)]   = (𝝿 /2)                    log(2e/(1+e) where x-0 to x=1 and y=0 to y=√(1-x^2)  *    In the integral ∬ (4-y) dydx, change the order of integration and evaluate the integral where x=2 to            x=4 and y=4/x to y=(20-4x)/(8-x)  *   Sketch the region of integration and evaluate ∬ xsiny dydx where x=0 to x=𝛑 and y-0 to...

PROBLEM :       Show that  ∫ xy 2 dy- x 2 y dx = 35 a 4 π /16 where C is the counter clockwise curve of the cardiod         r=a(1+cosθ ). SOLUTION :  Evaluate ∬ f(x,y) dxdy where f(x,y)= (2y-1)/ x+1 , and E is the region bounded by x=0,y=0, y=2x-        4.  *    Change the order of integration and hence show that ∬dx dy/[ ( 1+e^y) √(1-x^2-y^2)]   = (𝝿 /2)                    log(2e/(1+e) where x-0 to x=1 and y=0 to y=√(1-x^2)  *    In the integral ∬ (4-y) dydx, change the order of integration and evaluate the integral where x=2 to            x=4 and y=4/x to y=(20-4x)/(8-x)  *   Sketch the region of integration and evaluate ∬ xsiny dydx where x=0 to x=𝛑 and y-0 to y=x.  *   Sketch the region of integration and write an equivalent double integral with the order of ...

 PROBLEM :      Evaluate  ∬f(x,y)dxdy , where f(x,y)=x 2 +y 2 and E={(x,y)/y=x 2 ,x=2,y=1} SOLUTION :   *   Evaluate ∬ f(x,y) dxdy where f(x,y)= (2y-1)/ x+1 , and E is the region bounded by x=0,y=0, y=2x-        4.  *    Change the order of integration and hence show that ∬dx dy/[ ( 1+e^y) √(1-x^2-y^2)]   = (𝝿 /2)                    log(2e/(1+e) where x-0 to x=1 and y=0 to y=√(1-x^2)  *    In the integral ∬ (4-y) dydx, change the order of integration and evaluate the integral where x=2 to            x=4 and y=4/x to y=(20-4x)/(8-x)  *   Sketch the region of integration and evaluate ∬ xsiny dydx where x=0 to x=𝛑 and y-0 to y=x.  *   Sketch the region of integration and write an equivalent double integral with the order of integration      reversed for ∬ 3y ...