Question Number 101546 by mhmd last updated on 03/Jul/20

Commented bymhmd last updated on 03/Jul/20

help me sir

Answered by bobhans last updated on 03/Jul/20

(Q4) Area=2∫_0 ^3 9x^2 −x^4 dx =2(3x^3 −(1/5)x^5 )_0 ^3 = 2(81−((243)/5)) = 162(1−(3/5))= ((324)/5) ♥

Commented bymhmd last updated on 03/Jul/20

thank you sir can you help me in question one?

Answered by john santu last updated on 03/Jul/20

(Q3) (dy/(sin 2x)) = sec ((√y)) dx (dy/((√y) )) = sin 2x dx ⇒∫ y^(−(1/2)) dy = ∫ sin 2x dx 2(√y) = −(1/2) cos 2x + c (√y) = −(1/4) cos 2x + c ∴ y = (c −(1/4)cos 2x)^2 ⊛

Answered by abdomathmax last updated on 03/Jul/20

Q_2 ) ∫_0 ^2 (∫_0 ^(2π) (5ysin(5y)+xe^x )dydx =∫_0 ^2 A(x)dx with A(x) =∫_0 ^(2π) (5ysin(5y)+xe^x )dy A(x) =5 ∫_0 ^(2π) ysin(5y)dy +2πxe^x =5{ [−(y/5)cos(5y)]_0 ^(2π) +(1/5)∫_0 ^(2π) cos(5y)dy}+2πxe^x =5{−((2π)/5) +(1/(25))[sin(5y)]_0 ^(2π) } +2πxe^x =−2π +2πxe^x ⇒ I =∫_0 ^2 (−2π+2πxe^x )dx =−4π +2π ∫_0 ^2 xe^x dx =−4π +2π{ [xe^x ]_0 ^2 −∫_0 ^2 e^x dx} =−4π +2π{2e^2 −(e^2 −1)} =−4π +2π{e^2 +1) =−2π +2πe^2

Answered by john santu last updated on 04/Jul/20

(Q1) unit vector of A →e^� =((2i−4j+4k)/(√(36))) = (1/3)i−(2/3)j+(2/3)k →▽φ= 12xi +3zj +(3y−2z)k at (1,−2,1) ▽φ= 12i −6j −8k directional derivative → ∣e^� . ▽φ∣ = (1/3).(12)+((2/3)).(6)−(2/3)(8) =((12+12−16)/3) = (8/3) ⊛