Integral Calculus

  • ∫ -a a e ax cosnxdx

    see , you have to use by parts.. .. for simplification put “ nx “ = “t” … then dx = dt/n , and e ax = e (a/n)*t now , Take COSt as the first part and e (a/n)*t as the second part .. I =∫ -a a e ax...

    1 hour ago
  • 1/(sinx + secx)

    1/(sinx+secx) = cosx/(sinxcosx+1) = 2cosx/(2sinxcosx+2) =(cosx +sinx)/(3-(sinx-cosx) 2 ) + (cosx-sinx)/((cosx+sinx2) 2 +1) substitute sinx-cosx and t in first term and cosx-sinx as z in second term so...

    2 hours ago
  • integration of 1/1+x 4

    1/(1+x 4) = 1/2[(x 2 + 1)/(x 4 + 1) – (x 2 – 1)/(x 4 + 1)] ∫(x 2 + 1)/(x 4 + 1)dx = ∫(1 + 1/x 2 )/(x 2 + 1/x 2 )dx (dividing num & den by x 2 ) put x – 1/x 2 = t => x 2 + 1/x 2 = t 2 + 2 & 1 + 1/x 2...

    yesterday
  • ​Q. THE VALUE OF -

    Take either (cos x)^1/2 or (sin x)^1/2 as y and using substitution method, solve the eqn. Use the formula of dx to solve the eqn further. Putting the limits you’ll get your answer.You can even...

    3 days ago
  • b ∫ a e x dx

    Hi, It will be simply e b – e a...

    4 days ago
  • b ∫ a cos x dx

    2Cos[(a+b)/2] Sin[(a-b)/2] It would be great if you could post your doubt or the problem that you faced while solving the question. Possibly, your answer would not have been matching with the options...

    4 days ago
  • 2 ∫1 x 2 dx

    Ans:- 2 ∫ 1 x 2 dx =[(x 3 /3)] 1 2 =[(2 3 /3)-(1 3 /3)] =[(8/3)-(1/3)] =7/3

    4 days ago
  • pi/2 ∫ 0 sin x dx

    Ans:- pi/2 ∫ 0 sinx dx =[-cosx] 0 pi/2 =[-cos(pi/2)-(-cos0)] =[0+1] =1

    4 days ago
  • pi/2 ∫ 0 cos x dx

    Ans:- pi /2 ∫ 0 cosx dx -----–-- > ∫cosx dx=sinx =[sinx] 0 pi/2 =[sin(pi/2)-sin(0)] =[1-0] =1

    4 days ago
  • 4 ∫ 1 (3x 2 + 2x) dx

    Ans:- 4 ∫ 1 (3x 2 +2x)dx =[3(x 3 /3)+2(x 2 /2)] 1 4 =[x 3 +x 2 ] 1 4 =[(4 3 -1 3 )+(4 2 -1 2 )] =[(64-1)+(16-1)] =[63+15] =78

    4 days ago
  • 2 ∫ 0 (3x 2 – 2) dx

    Ans:- 2 ∫ 0 (3x 2 -2)dx =[3(x 3 /3)-2x] 0 2 =[x 3 -2x] 0 2 =[2 3 -2(2)] =8-4 =4

    4 days ago
  • 1 ∫ 0 (3x 2 + 5x) dx

    Ans:- 1 ∫ 0 (3x 2 +5x)dx =[3(x 3 /3)+5(x 2 /2)] 0 1 =[3(1/3)+5(1/2)] =[1+(5/2)] =(2+5)/2 =7/2

    4 days ago
  • 3 ∫ 0 ( x + 4) dx

    Ans:- 3 ∫ 0 (x+4)dx =[(x 2 /2)+4x] 0 3 =[(3 2 /2)+4(3)] =[(9/2)+12] =(9+24)/2 =33/2

    4 days ago
  • Definite Integration

    an integral expressed as the difference between the values of the integral at specified upper and lower limits of the independent variable

    4 days ago
  • 3 ∫ 1 (3x – 2) dx

    integration of 3x-2 from 1 to 3 is equal to (3(x^2)/2)-2x) from 1 to 3 =(3((9/2)-(1/2)))-(2(3-1)) =12-4 =8

    4 days ago
  • 2 ∫ 0 ( x + 3) dx

    integration of x+3 from 0 to 2 is equal to ((x^2)/2)+(3x) from 0 to 2 =(4/2)+(2*3)-0 =2+6 =8

    4 days ago
  • 1∫ -1 ( x + 3) dx

    Ans:- 1 ∫ -1 (x+3)dx =[(x 2 /2)+3x] -1 1 =[(1 2 )/2 -(-1 2 )/2]+3(1-(-1)) =(1/2)-(1/2)+6 =6

    4 days ago
  • 5 ∫ 0 (x + 1) dx

    Ans:- 5 ∫ 0 (x+1)dx =[(x 2 /2)+x] 0 5 =(25/2)+5 =(25+10)/2 =35/2

    4 days ago
  • 3 ∫ 1 ( 2x + 3) dx

    Ans:- 3 ∫ 1 (2x+3)dx =[2(x 2 /2)+3x] 1 3 =[3 2 -1 2 ]+3(3-1) =8+6 =14

    4 days ago
  • 5 ∫ 3 (2 – x) dx

    integration of 2-x from 3 to 5 is equal to (2x-((x^2)/2)) from 3 to 5 =(2*5-((5^2)/2))-(2*3-((3^2)/2)) =(10-(25/2))-(6-(9/2)) =(-5/2)-(3/2) =-8/2 =-4

    4 days ago
  • 2 ∫ 0 (x 2 + 1) dx

    Ans:- 2 ∫ 0 (x 2 +1)dx =[(x 2 +1 /2+1)] 0 2 =[(x 3 /3)+x] 0 2 =(8/3)+2 =(8+6)/3 =14/3

    4 days ago
  • 3 ∫ 2 (2x 2 + 1) dx

    integration of (2(X^2)+1)dx from 2 to 3 is equal to (2((x^3)/3)+x) from 2 to3 =(2((3^3)/3)+3)-(2((2^3)/3)+2) =(2*(27/3)+3)-(2*(8/3)+2) =(18+3)-((16/3)+2) =(21)-(22/3) =41/3

    4 days ago
  • 2 ∫ 1 (x 2 – 1) dx

    Ans:- 2 ∫ 1 (x 2 -1)dx=[(x 2+1 /2+1)-x] 1 2 =[(x 3 /3)-x] 1 2 =[(8/3)-(1/3)]-(2-1) =(7/3)-1 =(7-3)/3 =4/3

    4 days ago
  • 2 ∫ 0 (x 2 + 4) dx

    Ans:- 2 ∫ 0 (x 2 +4)dx=[(x 2+1 /2+1 )+4x] --------- since (x n+1 /n+1) [(x 3 /3)+4x] 0 2 =[(2 3 /3)+8]=[(8/3)+3]=(8+24)/3=32/3

    4 days ago
  • Q . PLEASE INTEGRATE THE FUNCTION WHOSE PIC IS ATTACHED.....

    thanks

    5 days ago
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