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*Image* ...
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A circular loop of radius R carries a current I.Another circular loop of radius r(<<R) carries a current i and is placed at the centre of the larger loop.The planes of the two circles are at right angle to each other.Fi ...
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evaluate *Image* answer given is π2. ...
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Integrate:: 4x3[d2/dx2(1-x2)5]dx from 0 to 1.Here d2/dx2 means the double derivative of the function w.r.t.x. This sum is meant for students only ,not for experts. ...
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Integrate:: 4x3[d2/dx2(1-x2)5]dx from 0 to 1.Here d2/dx2 means the double derivative of the function w.r.t.x. This sum is meant for students only ,not for experts. ...
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If x>0 then integrate (2x-[2x])2 from 0 to 100[x] where [.] means G.I.F. ...
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1) Prove that ∫x3 2ax-x2 dx=7πa5/8. 2) Integrate: [ x4/(1-x4)] *cos-1(2x/(1+x2). Upper limit=1/ 3 ;Lower limit=-1/ 3 . 3) Integrate::: 2-x2/[(1+x) 1-x2 ].Upper limit=1;Lower limit=0. 4) Integrate:::: log(1+tanx) from 0 to ...
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1) Prove that ∫x3 2ax-x2 dx=7πa5/8. 2) Integrate: [ x4/(1-x4)] *cos-1(2x/(1+x2). Upper limit=1/ 3 ;Lower limit=-1/ 3 . 3) Integrate::: 2-x2/[(1+x) 1-x2 ].Upper limit=1;Lower limit=0. 4) Integrate:::: log(1+tanx) from 0 to ...
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1) Prove that ∫x3 2ax-x2 dx=7πa5/8. 2) Integrate: [ x4/(1-x4)] *cos-1(2x/(1+x2). Upper limit=1/ 3 ;Lower limit=-1/ 3 . 3) Integrate::: 2-x2/[(1+x) 1-x2 ].Upper limit=1;Lower limit=0. 4) Integrate:::: log(1+tanx) from 0 to ...
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1) Prove that ∫x3 2ax-x2 dx=7πa5/8. 2) Integrate: [ x4/(1-x4)] *cos-1(2x/(1+x2). Upper limit=1/ 3 ;Lower limit=-1/ 3 . 3) Integrate::: 2-x2/[(1+x) 1-x2 ].Upper limit=1;Lower limit=0. 4) Integrate:::: log(1+tanx) from 0 to ...
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INTEGRATE: 1) (x3+3)/x6(x2+1). Upper limit=∞;Lower limit=0 2) xe-x/( 1-e-2x ). Upper limit=0;Lower limit=∞. 3) If f(x) be a function satisfying f'(x)=f(x) with f((0)=1 and g be the function satisfying f(x) + g(x)=x2,then ...
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INTEGRATE: 1) (x3+3)/x6(x2+1). Upper limit=∞;Lower limit=0 2) xe-x/( 1-e-2x ). Upper limit=0;Lower limit=∞. 3) If f(x) be a function satisfying f'(x)=f(x) with f((0)=1 and g be the function satisfying f(x) + g(x)=x2,then ...
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Integrate: 1) (1+x-2/3)/(1+x) 2) cos2x /sinx 3)(x2 + n(n-1))/(xsinx+ncosx)2 ...
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Integrate: 1) (1+x-2/3)/(1+x) 2) cos2x /sinx 3)(x2 + n(n-1))/(xsinx+ncosx)2 ...
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Integrate: 1) (1+x-2/3)/(1+x) 2) cos2x /sinx 3)(x2 + n(n-1))/(xsinx+ncosx)2 ...
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Integrate: 1) [(x-1) x4+2x3-x2+2x+1 ]/x2(x+1) 2) (x2-1)/(x3 2x4-2x2+1 ) ...
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Integrate: 1) [(x-1) x4+2x3-x2+2x+1 ]/x2(x+1) 2) (x2-1)/(x3 2x4-2x2+1 ) ...
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Integrate: 1) [(x-1) x4+2x3-x2+2x+1 ]/x2(x+1) 2) (x2-1)/(x3 2x4-2x2+1 ) ...
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Why is pyridine used in the reaction of R-OH + PCl3 to make R-Cl? ...
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try this integration: \frac{x^{2}-2}{x^{3}\sqrt{x^{2}-1}} ...
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INTEGRATE: emsin-1x ...
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If f(x) and φ(x) are continuous function in[0,4] satisfying f(x)=f(4-x),φ(x) + φ(4-x)=3 and ∫40f(x)dx=2,then the value of ∫40 f(x)φ(x)dx is: NOTE:The upper and lower limits are 4 and 0 ...
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*Image* ...
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If each ai>0, then the shortest distance between the point(0,-3) and the curve y=1+a1x2 + a2x4 + ...........................................+anx2n is a) 1 b) 2 c) 3 d) 4 ...
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what is the mechanism for the intramolecular wurtz reaction? ...
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*Image* ...
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Find the acceleration of the block M. Given that the coefficient of friction between the blocks M and m is u1 and the coefficient of friction between the block M and the floor is u2. Clearly show your calculations. *Image* ...
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Find the acceleration of the block M. Given that the coefficient of friction between the blocks M and m is u1 and the coefficient of friction between the block M and the floor is u2. Clearly show your calculations. *Image* ...
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a battrry of emf 10V and internal resistance 0.5 ohms is connected across a variable resistance R. What is the value of R for which the power delivered in it is maximum? ...
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A rod of length 10 m. is travelling on a smooth,level,horizontal surface.Beyond a point A,the floor ceases to be smooth and the coefficient of friction is k=0.1.The rod is travelling such that it is moving towards the point A ...
